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MECHANICS OF ENGINEERING. 



COMPRISING 

STATICS AND DYNAMICS OF SOLIDS; AND THE 
MECHANICS OF THE MATERIALS OF CON- 
STRUCTIONS, OR STRENGTH AND 
ELASTICITY OF BEAMS, COL- 
UMNS, ARCHES, SHAFTS, 
ETC. 



FOR USE IN TECHNICAL SCHOOLS. 

A 

IRVING P. CHURCH, C.E., 



3/ c 



ASSISTANT PROFESSOR OF CIVIL ENGINEERING, CORNELL UNIVERSITY. 

(In charge of Applied Mechanics.) 




NEW YORK: 

JOHN WILEY & SONS, 

15 Astor Place. 

1888. 



f>,Jo 



1*c$ 






Copyright, 1888, 
Bv IRVING P. CHURCH, 



£-11 6 9 



/-if 



PREFACE. 



For the engineering student, pursuing the study of Applied Mechan- 
ics as part of his professional training, and not as additional mathe- 
mathical culture, not only is a thoroughly x systematic, clear, and 
consistent treatment of the subject quite essential, but one which pre- 
sents the quantities and conceptions involved in as practical and con- 
crete a form as possible, with all the aids of the printer's and engraver's 
arts ; and especially one which, besides showing the derivation of 
formulae from principles, illustrates, inculcates, and lays stress on 
correct numerical substitution and the consistent and proper use of 
units of measurement; for without this no reliable results can be 
reached, and the principal object of these formulae is frustrated. 

With these requirements in view, and aided by the experience of ten 
years in teaching the Mechanics of Engineering at this institution, the 
writer has been led to prepare the present work, in which attention is 
called to the following features : 

The diagrams are very numerous (about one to every page ; an appeal 
to the eye is often worth a page of verbal description). 

The symbols for distances, angles, forces, etc. , used in the algebraic 
work are, as far as possible, inserted directly in the diagrams, render- 
ing the latter full and explicit, and thus saving time and mental effort 
to the student. In problems in Dynamics three kinds of arrows are 
used to distinguish forces, velocities, and accelerations, respectively, 
and thus to prevent confusion of ideas. 

Illustrations and examples of a practical nature, both algebraic and 
numerical, are of frequent occurrence. 

Formulae are divided into two classes ; those (homogeneous) admit- 
ting of the use of any system of units whatever for measurements of 
force, space, mass, and time, in numerical substitution; and those 
which are true for specified units only. Attention is repeatedly di- 
rected to the matter of correct numerical substitution, especially in 
Dynamics, where time and mass, as well as force and space, are among 
the quantities considered. The importance, in this connection, of 
frequent mention of the quality of the various kinds of quantity em- 
ployed, is also recognized, and a corresponding phraseology adopted. 

The definition of force (§ 3) is made to include and illustrate Newton's 
law of action and reaction, the misconception of which leads to such 
lengthy discussions in technical journals every few years. 

In the matter of "Centrifugal force," the artificial method, so com- 
monly adopted, of regarding a particle moving uniformly m a circle 
as in equilibrium, i. e. , acted on by a balanced system of forces, one of 
which is the "Centrifugal force," has been avoided, as being at vari- 
ance with a system of Mechanics founded on Newton's laws, according 
to the first of which a particle moving in any other than a straight line 
cannot be in equilibrium. In such a system of Mechanics nothing can 



iv PREFACE. 

\>e recognized as a force which is not a definite pull, push, pressure, 
rub, attraction or repulsion, of one body upon, or against, another. 

It is true that the artificial nature of the method referred to is in 
some text-books fully explained in the context, (in Goodeve's Steam 
Engine, for instance, in treating the governor ball,) but is too often not 
mentioned at all, so that the student risks being led into error in 
attempting kindred problems by what would then seem to him correct 
methods. 

The general theorem of Work and Energy in machines is developed 
gradually by definite and limited steps, in preference to giving a single 
demonstration which, from its generality, might be too vague and ab- 
struse to be readily grasped by the student. 

In the use of the Calculus, (in the elements of which the student is 
supposed to have had the training usually given in technical schools by 
the end of the second year) the integral sign is always used to indicate 
summation (except on p. 357) while the name of anti-derivative of a 
given function (of one variable) has been adopted for that function whose 
derivative, or differential co-efficient, is the given function (see §253. ) 

The signs -j and || are used for perpendicular and parallel, respect- 
ively. 

In Torsion and Flexure of Beams, the well worn and simple theories 
of Navier have been thought sufficient for establishing practical for- 
mulae for safe loads and deflections of beams and shafts ; and promi- 
nence has been given to the methods of designing the cross-sections 
and riveting of built-beams and plate-girders, forming the basis of the 
tables and rules usually given in the pocket-books of our iron and steel 
manufacturers. 

The analytical treatment of the continuous girder is not presented in 
the general case, preference being given to the graphic method by 
Mohr, as greatly superior in simplicity, directness, and interest. For 
Bimilar reasons the graphics of the arch of masonry is to be preferred to 
the analytical chapter on Linear Arches, whose insertion is chiefly a 
concession to the mathematical student, as are also §§ 119, 198, 234, 235, 
264, 265, 266, 284, 287, 291, and 297. 

The graphics of curved beams or arch ribs is made to precede that of 
the straight girder, since the treatment of the latter as a particular case 
of the former is then a comparatively simple matter. Hence Prof. 
Eddy's methods* (inserted by his kind permission) for the arch rib of 
hinged ends, and also that of fixed ends, are presented as special geo- 
metrical devices, instead of being based on Prof. Eddy's general theorem 
(involving a straight girder of the same section and mode of support). 

Acknowledgment is also due Prof. Burr and Prof. Robinson, for 
their cordial consent to the use of certain items and passages from 
their works; (see §§206, 212, 220, and 297.) 

Cornell University, Ithaca, N. Y., Sept., 1887. 

* See pp. 14 and 35 of " Researches in Graphical Statics." by Prof. H. T. Eddy, 
C.E., Ph. D.; published by D. Van Nostrand, New York, 1878 : reprinted from Van Nos- 
trand's Magazine for 1877. 



TABLE OF CONTENTS. 



PRELIMINARY CHAPTER. 

PAGE 

§§ l-15a. Definitions. Kinds of Quantity. Homogeneous Equa- 
tions. Parallelogram of Forces. 1 

PAET I. STATICS. 

Chapter I. Statics of a Material Point. 
§§ 16-19. Composition and Equilibrium of Concurrent Forces 8 

Chapter II. Parallel Forces and the Centre op Gravity. 

§§20-22. Parallel Forces 13 

§§ 23-276. Centre of Gravity. Problems. Centrobaric Method. . . 1& 

Chapter III. Statics op a Rigid Body. 

§§28-34. Couples 27 

§§ 35-39. Composition and Equilibrium of Non-concurrent Forces. 31 

Chapter IV. Statics of Flexible Cords. 

§§ 40-48. Postulates. Suspended Weights. Parabolic Cord. Cat- 
enary 42 

PART II. DYNAMICS. 
Chapter I. Rectilinear Motion of a Material Point. 

§§ 49-55. Uniform Motion. Falling Bodies. Newton's Laws. 

Mass 49 

§§ 56-60. Uniformly Accelerated Motion. Graphic Representa- 

tions. Variably Accelerated Motions. Impact. . . 54 

Chapter II. Virtual Velocities. 

§§ 61-69. Definitions and Propositions. Connecting-rod. Prob- 
lems 6? 



Yl CONTENTS. 

Chapter III. Curvilinear Motion op a Material Point. 

• PAGE 

§§ 70-74. Composition of Motions, of Velocities, etc. General 

Equations 72 

§§ 75-84. Normal Acceleration. Centripetal and Centrifugal 
Forces. Simple Pendulums. Projectiles. Rela- 
tive Motion 77 

Chapter IV. Moment of Inertia. 

§§ 85-94. Plane Figures. Rigid Bodies. Reduction Formulae. 

The Rectangle, Triangle, etc. Compound Plane 

Figures. Polar Moment of Inertia 91 

§§ 95-104. Rod . Thin Plates. Geometric Solids 98 

§§ 105-107. Numerical Substitution. Ellipsoid of Inertia 102 

Chapter V. Dynamics of a Rigid Body. 

§§ 108-115. Translation. Rotation about a Fixed Axis. Centre of 

Percussion 105 

§§ 116-121. Torsion Balance. Compound Pendulum. The Fly- 
wheel 116 

§§122-123. Uniform Rotation. " Centrifugal Action." Free Axes. 125 

§§ 124-126. Rolling Motions. Parallel Rod of Locomotive 130 

Chapter VI. Work, Energy, and Power. 

§§ 127-134. Work. Power. Horse-power. Kinetic Energy 133 

§§135-138. Steam-hammer. Pile-driving. Inelastic Impact 138 

§§ 139-141. Rotary Motion. Equivalent Systems of Forces. Any 

Motion of a Rigid Body 143 

§§ 142-146. Work and Kinetic Energy in a Moving Machine of 

Rigid Parts 147 

§§ 147-155. Power of Motors. Potential Energy. Heat, etc. Dy- 
namometers. Atwood's Machine. Boat-rowing. 
Examples 153 

Chapter VII. Friction. 

§§ 156-164. Sliding Friction. Its Laws. Bent Lever 164 

§§ 165-171. Ax'ie - friction. Friction Wheels. Pivots. Belting. 

Transmission of Power by Belting 175 

§§ 172-177. Rolling Friction. Brakes. Engine-friction. Anoma- 
lies in Friction. Rigidity of Cordage . Examples. 186 



CONTENTS. VU 

PAKT III. STRENGTH OF MATEKIALS. 

(OR MECHANICS OF MATEKIALS.) 

CHAPTER I. ELEMENTARY STRESSES AND STRAINS. 

§§ 178-182. Stress and Strain ; of Two Kinds. Oblique Section 

of Rod in Tension 195 

§§ 182a-190. Hooke's Law. Elasticity. Safe Limit. Elastic 
Limit. Rupture. Modulus of Elasticity. 
Isotropes. Resilience. Internal Stress. Tem- 
perature Stresses 201 

TENSION. 

§§ 191-195. Hooke's Law by Experiment Strain Diagrams. 

.Lateral Contraction. Modulus of Tenacity 207 

§§ 196-199. Resilience of Stretched Prism. Load Applied Sud- 
denly. Prism Under Its Own Weight. Solid 
of Uniform Strength. Temperature Stresses . . 213 

COMPRESSION OP SHORT BLOCKS. 

§§ 200-202. Short and Long Columns. Remarks on Crushing. . . 218 

EXAMPLES IN TENSION AND COMPRESSION. 

§§ 203-206. Tables. Examples. Factor of Safety. Practical 

Notes , 220 

SHEARING. 

§§ 207-213. Rivets. Shearing Distortion. Table. Punching. 

Examples ' 225 

CHAPTER II. TORSION. 

§§ 214-220. Angle and Moment of Torsion. Torsional Strength, 

Stiffness, and Resilience. Non-Circular Shafts 233 

§§ 221-223. Transmission of Power. Autographic Testing Ma- 
chine. Examples 238 

CHAPTER III. FLEXURE OF HOMOGENEOUS PRISMS UNDER 
PERPENDICULAR FORCES IN ONE PLANE. 

§§ 224-232a. The Common Theory. Elastic Forces. Neutral 
Axis. The "Shear" and "Moment." Flex- 
ural Strength and Stiffness. Radius of Curva- 
ture. Resilience 244 

ELASTIC CURVES. 

§§ 233-238. Single Central Load ; at Rest, and Applied Suddenly. 

Eccentric Load. Uniform Load. Cantilever. . 252 



Ylll 



CONTENTS. 



SAFE LOADS. 

§§ 239-246. Maximum Moment. Shear = x-Derivative of the 
Moment. Simple Beams With Various Loads. 
Comparative Strength and Stiffness of Rectan- 
gular Beams 262 

§§247-252. Moments of Inertia. " Shape Iron." I-Beams. Etc. 

Cantilevers. Tables. Numerical Examples. .. 273 

SHEARING STRESSES IN FLEXURE. 

§§ 253-257. Shearing Stress Parallel to Neutral Surface ; and in 
Cross Section. "Web of I-Beam. Riveting of 
Built Beams 284 



SPECIA.L PROBLEMS IN FLEXURE. 

§§ 258-265. Designing Sections of Built Beams. Moving Loads. 
Special Cases of Quiescent Loads. Hydrostatic 
Load 

§§ 266-270. Derivatives of Ordinate of Elastic Curve. Weight 
Falling on Beam. Crank Shaft. Other Shafts. 



295 



31fr 



CHAPTER IV. FLEXURE; CONTINUED. 

CONTINUOUS GIRDERS. 

§§ 271-276. Analytical Treatment of Symmetrical Cases of Beams 

on Three Supports ; also, Built in ; (see § 391.). . 320 

THE DANGEROUS SECTION IN NON-PRISMATIC BEAMS. 

§§ 277-279. Double Truncated Wedge, Pyramid, and Cone 333 

NON-PRISMATIC BEAMS OF UNIFORM STRENGTH. 

§§ 280-289. Parabolic and Wedge-Shaped Beams. I-Beam. 

Elliptical Beam and Cantilevers 335 

DEFLECTION OF BEAMS OF UNIFORM STRENGTH. 

§§ 290-293. Parabolic and Wedge-Shaped Beams. Special Prob- 
lems 342; 



CHAPTER V. FLEXURE OF PRISMATIC BEAMS UNDER 
OBLIQUE FORCES. 

3 294-301. Gravity and Neutral Axes. Shear, Thrust, and 
Stress-Couple. Strength and Stiffness of Ob- 
lique Cantilever. Hooks. Crane 34T 



CONTENTS. IX 

CHAPTER VI. FLEXURE OF LONG COLUMNS. 

§§ 302-308. End-Conditions. Euler's Formula. "Incipient 
Flexure." Hodgkinson's Formulae. Rankine's 
Formula 363 

§§ 309-314. Radii of Gyration. Built Columns. Trussed Gird- 
ers. Buckling of Web-Plates. Examples 376 

CHAPTER VII. LINEAR ARCHES (OF BLOCKWORK.) 

§§ 315-323. Inverted Catenary. Parabolic Arch. Circular Arch. 

Transformed Catenary as Arch 386 

CHAPTER VIII. ELEMENTS OF GRAPHICAL STATICS. 

§§ 321-326. Force Polygons. Concurrent and Non-Concurrent 
Forces in a Plane. Force Diagrams. Equili- 
brium Polygons 397 

§§ 327-332. Constructions for Resultant, Pier-Reactions, and 
Stresses in Roof Truss. Bow's Notation. The 
Special Equilibrium Polygon 402 

CHAPTER IX. GRAPHICAL STATICS OF VERTICAL FORCES. 

§§ 333-336. Jointed Rods. Centre of Gravity 412 

§§ 337-343. Useful Relations Between Force Diagrams and Then- 
Equilibrium Polygons 415 

CHAPTER X. RIGHT ARCHES OF MASONRY. 

§§344-352. Definitions. Mortar and Friction. Pressure in Joints. 
Conditions of Safe Equilibrium. True Linear 

Arch 421 

§§ 353-357. Arrangement of Data for Graphic Treatment 428 

§§ 358-362. Graphical Treatment of Arch. Symmetrical and 

Unsymmetrical Cases 431 

CHAPTER XI. ARCH-RIBS. 

§§ 334-374. Mode of Support. Special Equilibrium Polygon and 
its Force Diagram. Change in Angle Between 
Rib Tangents. Displacement of Any Point on 
Rib 438 

§§ 374o-378a. Graphical Arithmetic. Summation of Products. 
Moment of Inertia by Graphics. Classification 
of Arch-Ribs 450 

§§ 379-388. Prof. Eddy's Graphical Method for Arch-Ribs of 
Hinged Ends ; and of Fixed Ends. Stress 
Diagrams. Temperature Stresses. Braced 
Arches 461 



X CONTENTS. 

HORIZONTAL STRAIGHT GIRDERS. 

§§ 389-390. Prismatic Girder Supported at the Extremities, also 

with Ends Fixed Horizontally 471 



CHAPTER XII. GRAPHICS OF CONTINUOUS GIRDERS. 

§§ 391-397. The Elastic Curve an Equilibrium Polygon. Mohr's 
Theorem. Example and Numerical Case. End- 
Tangents. Re-arrangement of Moment- Area. . 484 

§§ 398-405. Positive Moment- Areas ; Amount and Gravity Ver- 
tical. Construction of the "False Polygons" 
in Any Given Case. Moment Curves, Shears, 
and Reactions. Numerical Example in Detail. 
Remarks 497 



MECHANICS OF ENGINEERING. 



PRELIMINARY CHAPTER. 

1. Mechanics treats of the mutual actions and relative mo- 
tions of material -bodies, solid, liquid, and gaseous; and by 
'Mechanics of Engineering is meant a presentment of those 
principles of pure mechanics, and their applications, which are 
of special service in engineering problems. 

2. Kinds of Quantity. — Mechanics involves the following 
fundamental kinds of quantity: Space, of one, two, or three 
dimensions, i.e., length, surface, or volume, respectively ; time, 
which needs no definition here; force and mass, as defined be- 
low; and abstract numbers, whose values are independent of 
arbitrary units, as, for example, a ratio. 

3. Force. — A force is one of a pair of equal, opposite, and 
simultaneous actions between two bodies, by which the state 
of their motions is altered or a change of form in the bodies 
themselves is effected. Pressure, attraction, repulsion, and 
traction are instances in point. Muscular sensation conveys 
the idea of force, while a spring-balance gives an absolute 
measure of it, a beam-balance only a relative measure. In 
accordance with Newton's third law of motion, that action and 
reaction are equal, opposite, and simultaneous, forces always 
occur in pairs; thus, if a pressure of 40 lbs. exists between 
bodies A and B, if A is considered by itself (i.e., "free"), 
apart from all other bodies whose actions upon it are called 
forces, among these forces will be one of 40 lbs. directed from 
B toward A. Similarly, if B is under consideration, a force 



2 MECHANICS OF ENGINEERING. 

of 40 lbs. directed from A toward B takes its place among the 
forces acting on B. This is the interpretation of Newton's 
third law. 

In conceiving of a force as applied at a certain point of a 
body it is useful to imagine one end of an imponderable spiral 
spring in a state of compression (or tension) as attached at the 
given point, the axis of the spring having the given direction 
of the force. 

4. Mass is the quantity of matter in a body. The masses of 
several bodies being proportional to their weights at the same 
locality on the earth's surface, in physics the weight is taken 
as the mass, but in practical engineering another mode is used 
for measuring it (as explained in a subsequent chapter), viz.: 
the mass of a body is equal to its weight divided by the ac- 
celeration of gravity in the locality where the weight is taken, 
or, symbolically, M— G -~ g. This quotient is a constant 
quantity, as it should be, since the mass of a body is invariable 
wherever the body be carried. 

5. Derived Quantities. — All kinds of quantity besides the 
fundamental ones just mentioned are compounds of the latter, 
formed by multiplication or division, such as velocity, accele- 
ration, momentum, work, energy, moment, power, and force- 
distribution. Some of these are merely names given for 
convenience to certain combinations of factors which come 
together not in dealing with first principles, but as a result of 
common algebraic transformations. 

6. Homogeneous Equations are those of such a form that they 

are true for any arbitrary system of units, and in which all 

terms combined by algebraic addition are of the same kind. 

qf 
Thus, the equation s — ~ (in which g = the acceleration of 

gravity and t the time of vertical fall of a body in vacuo, 
from rest) will give the distance fallen through, s, whatever 
units be adopted for measuring time and distance. But if for 



PRELIMINARY CHAPTER. 3 

•g we write the numerical value 32.2, which it assumes when 
time is measured in seconds and distance in feet, the equation 
s = lQ.lf is true for those units alone, and the equation is not 
of homogeneous form. Algebraic combination of homogeneous 
equations should always produce homogeneous equations ; if 
not, some error has been made in the algebraic work. If any 
equation derived or proposed for practical use is not homogene- 
ous, an explicit statement should be made in the context as to 
the proper units to be employed. 

7. Heaviness. — By heaviness of a substance is meant the 
weight of a cubic unit of the substance. E.g. the heaviness of 
fresh water is 62.5, in case the unit of force is the pound, 
and the foot the unit of space; i.e., a cubic foot of fresh 
water weighs 62.5 lbs. In case the substance is not uniform 
in composition, the heaviness varies from point to point. If 
the weight of a homogeneous body be denoted by 6r, its volume 
by V y and the heaviness of its substance by y, then G = Vy. 



Weight in Pounds of a Cubic Foot (i.e., the heaviness) of various 

Materials. 



Anthracite, solid 100 

broken 57 

Brick, common hard 125 

soft 100 

Brick-work, common 112 

Concrete 125 

Earth, loose 72 

as mud 102 

Granite 164 to 172 

Ice 58 

Iron, cast 450 

" wrought 480 



Masonry, dry rubble 138 

" dressed granite or 

limestone 165 

Mortar 100 

Petroleum 55 

Snow 7 

" wet 15 to 50 

Steel 490 

Timber 25 to 60 

Water, fresh 62 . 5 

sea 64.0 



8. Specific Gravity is the ratio of the heaviness of a material 
to that of water, and is therefore an abstract number. 

9. A Material Point is a solid body, or small particle, whose 
-dimensions are practically nothing, compared with its range of 
motion. 



4 MECHANICS OF ENGINEERING. 

10. A Rigid Body is a solid, whose distortion or change of 
form under any system of forces to be brought upon it in 
practice is, for certain purposes, insensible. 

11. Equilibrium. — When a system of forces applied to a 
body produces the same effect as if no force acted, so far as 
the state of motion of the body is concerned, they are said to- 
be balanced, or to be in equilibrium. 

12. Division of the Subject. — /Statics will treat of bodies at 
rest, i.e., of balanced forces or equilibrium; dynamics, of 
bodies in motion ; strength of materials will treat of the effect 
of forces in distorting bodies ; hydraulics, of the mechanics- 
of liquids ; pneumatics, of the mechanics of gases. 

13. Parallelogram of Forces. — Duchayla's Proof. To fully 
determine a force we must have given its amount, its direc- 
tion, and its point of application in the body. It is generally 
denoted in diagrams by an arrow. It is a matter of experience 
that besides the point of application already spoken of any 
other may be chosen in the line of action of the force. Tin's 
is called the transmissibility of force; i.e., so far as the state of 
motion of the body is concerned, a force may be applied any- 
where in its line of action. 

The Resultant of two forces (called its components) applied 
at a point of a body is a single force applied at the same point y 
which will replace them. To prove that this resultant is given 
in amount and position by the diagonal of the parallelogram 
formed on the two given forces (conceived as laid off to some 
scale, so many pounds to the inch, say), Duchayla's method 
requires four postulates, viz. : (1) the resultant of two forces 
must lie in the same plane with them ; (2) the resultant of two 
equal forces must bisect the angle between them ; (3) if one of 
the two forces be increased, the angle between the other force 
and the resultant will be greater than before; and (4) the trans 
missibility of force, already mentioned. Granting these, we- 
proceed as follows (Fig. 1) : Given the two forces P and Q = 



PRELIMINARY CHAPTER. 5 

r + P" {P' and P" being each equal to P, so that Q = 2P), 
applied at 0. Transmit P" to A. Draw the parallelograms 
OB and AD ; OD will also be a parallelogram. By postulate 
■(2), since OB is a rhombus, P and P / at may be replaced by 
a single force R' acting through B. Transmit R' to B and 
replace it by P and P ' . Transmit P from P to A, P' from 
P to D. Similarly P and P", at A, may be replaced by a 
single force P" passing through D ; transmit it there and re- 
solve it into P and P" . P' is already at D. Hence P and 
P' + P", acting at P, are equivalent to P and P -f P" act- 
ing at 0, in their respective directions. Therefore the result- 
ant of P and P' -fr P" must lie in the line OD, the diagonal 
•of the parallelogram formed on P and Q = 2P at 0. Similarly 





Q' P" o 


%- x 


C/ F] /B 







H\E 

P / %/ Rri d >*r 

Fig. 1. Fig. 2. 

this may be proved (that the diagonal gives the direction of 
the resultant) for any two forces P and mP ; and for any two 
forces nP and mP, m and n being any two whole numbers, 
i.e., for any two commensurable forces. When the forces are 
incommensurable (Fig. 2), P and Q being the given forces, 
we may use a reduotio ad ahsurdum, thus : Form the parallelo- 
gram OD on P and Q applied at 0. Suppose for an instant 
that R the resultant of P and Q does not follow the diagonal 
OD, but some other direction, as OD' . Note the intersection 
P, and draw HC parallel to DB. Divide P into equal parts, 
each less than HD ; then in laying off parts equal to these from 
O along OB, a point of division will come at some point F 
between C and B. Complete the parallelogram OFEG. The 
force Q" = OF is commensurable with P, and hence their 



XR 




F\ 




\o 


Q, 


/ \ 




\^B 


xD/ 


P 


R* 


Fig. 3. 





D MECHANICS OF ENGINEERING. 

resultant acts along OK Now Q is greater than Q", while P 
makes a less angle with P than OE, which is contrary to pos- 
tulate (3); therefore P cannot lie outside of the line OD. 
Q. E. D. * 

It still remains to prove that the resultant is represented in 
amount, as well as position, by the diagonal. OD (Fig. 3) is 
the direction of P the resultant of P and 
Q\ required its amount. If P' be a force 
equal and opposite to P it will balance P 
and Q ; i.e, the resultant of P' and P 
must lie in the line QO prolonged (besides 
being equal to Q). We can therefore de- 
termine P' by drawing BA parallel to DO to intersect QO 
prolonged in A ; and then complete the parallelogram on 
PA and BO. Since OFAB is a parallelogram P must =BA, 
and since OABD is a parallelogram BA=OD; therefore 
W=OD and also P= OP. Q. E. D. 

Corollary. — The resultant of three forces applied at the same 
point is the diagonal of the parallelopiped formed on the three 
forces. 

14. Concurrent forces are those whose lines of action intersect 
in a common point, while non-concurrent forces are those which 
do not so intersect ; results obtained for a system of concurrent 
forces are really derivable, as particular cases, from those per- 
taining to a system of non-concurrent forces. 

15. Resultant. — A single force, the action of which, as re- 
gards the state of motion of the body acted on, is equivalent to 
that of a number of forces forming a system, is said to be the 
Resultant of that s} r stem, and may replace the system ; and con- 
versely a force which is equal and opposite to the resultant of 
a system will balance that system, or, in other words, when it 
is combined with that system there will result a new system in 
equilibrium. 

In general, as will be seen, a given system of forces can al- 



PRELIMINARY CHAPTER. 7 

ways De replaced by two single forces, but these two can be 
combined into a single resultant only in particular cases. 

15a. Equivalent Systems are those which may be replaced by 
the same set of two single forces — or, in other words, those 
which have the same effect, as to state of motion, upon the 
given body. 

15b. Formulae. — If in Fig. 3 the forces P and Q and the angle a = 
PO Q are given, we have, for the resultant, 



R = D = V P 2 + Q 2 + 2 P Q cos a. 

(If a is > 90° its cosine is negative.) In general, given any three parts 
of either plane triangle OB Q, or D B, the other three may be obtained 
by ordinary trigonometry. Evidently if a = 0, B = P + Q ; if a = 
180°, B = P- Q ; and if a = 90°, B = V P 2 + Q\ 

15c. Varieties of Forces. — Great care should be used in deciding 
what may properly be called forces. The latter may be divided into ac- 
tions by contact, and actions at a distance. If pressure exists between two 
bodies and they are perfectly smooth at the surface of contact, the pressure 
(or thrust, or compressive action), of one against the other constitutes a force, 
whose direction is normal to the tangent plane at any point of contact (a 
matter of experience) ; while if those surfaces are not smooth there may also 
exist mutual tangential actions or friction. (If the bodies really form a 
continuous substance at the surface considered, these tangential actions are 
called shearing forces.) Again, when a rod or wire is subjected to tension, 
any portion of it is said to exert a pull or tensile force upon the remainder ; 
the ability to do this depends on the property of cohesion. The foregoing 
are examples of actions by contact. 

Actions at a distance are exemplified in the mysterious attractions, or re- 
pulsions, observable in the phenomena of gravitation, electricity, and mag- 
netism, where the bodies concerned are not necessarily in contact. By the 
term weight we shall always mean the force of the earth's attraction on the 
body in question, and not the amount of matter in it. 

[Note.— In some common phrases, such as " The tremendous force " of a heavy body in 
rapid motion, the word force is not used in a technical sense, but signifies energy (as ex- 
plained in Chap. VI.). The mere fact that a body is in motion, whatever its mass and 
velocity, does not imply that it is under the action of any force, necessarily. For instance, 
at any point in the path of a cannon ball through the air, the only forces acting on it.are 
the resistance of the air and the attraction of the earth, the latter having a vertical and 
downward direction.] 



PART I.-STATICS. 



CHAPTER I. 

STATICS OF A MATERIAL POINT. 

16. Composition of Concurrent Forces. — A system of forces 
acting on a material point is necessarily composed of concurrent 
forces. 

Case I. — All the forces in One Plane. Let be the 
material point, the common point of application of all the 
forces ; P„ P„ etc., the given forces, making 
—ffi angles a l9 a„ etc., with the axis X. By the 

■■/A —jpfl parallelogram of forces P x may be resolved 

•Js^HQ I into and replaced by its components, i^cos a 

- 1 -* — l — * acting along X, and P x sin a along Y. 

FlG - 4 - Similarly all the remaining forces may be re- 

placed by their Xand Y components. We have now a new 
system, the equivalent of that first given, consisting of a set of 
X forces, having the same line of application (axis X), and a 
set of Y forces, all acting in the line Y. The resultant of the 
X forces being their algebraic sum (denoted by 2X) (since 
they have the same line of application) we have 

2X = P 1 cos a 1 -f- P t cos a 2 -f- etc. = 2(P cos <z), 

and similarly 

2Y=P t sin a 1 -j- P 2 sin a t -f- etc. = 2(P sin a). 

These two forces, 2X and 2 Y, may be combined by the 
parallelogram of forces, giving P = \/(^XY -f- (2 Y)* as tne 
single resultant of the whole system, and its direction is deter- 

2Y 

mined by the angle a-, thus, tan a = -^rp-; see Fig. 5. For 

equilibrium to exist, P must = 0, which requires, separately. 



STATICS OF A MATERIAL POINT. 



" 9 



2X=0, and 2Y = (for the two squares (2J) 2 and 
(2 Y) 2 can neither of them be negative quantities). 

Case II. — The forces having any directions in space, 
but all applied at O, the material point. Let P 19 P„ 
etc., be the given forces, P x making the angles a„ j3^ and y x , 
respectively, with three arbitrary axes, X, Y, and Z (Fig. 6), 
at right angles to each other and intersecting at 0, the origin. 
Similarly let <* 2 , /? 2 , y„ be the angles made by P 2 with these 
axes, and so on for all the forces. By the parallelopiped of 
forces, P x may be replaced by its components. 

X x = P x cos a x , Y 1 = P x cos j3 x , and Z x = P x cos y x ; and 






Fig. 6. 



Fig. 7. 



similarly for all the forces, so that the entire system is now 
replaced by the three forces, 

2X = P x cos a x -\- P* cos a 2 + etc ; 
2Y = P x cos A + P 2 cos p % + etc; 
2Z = P x cos y x + P 2 cos y 2 -\- etc ; 

and finally by the single resultant 

R = V(zxy + {2 Yy + (szy. 

Therefore, for equilibrium we must have separately, 
2X= 0, 2Y = 0, and 2Z = 0. 
i?'s position may be determined by its direction cosines, viz., 



2X A 

cos a = — p- ; cos p 



2Y 



2Z 



7r> G0Sr = -p 



17. Conditions of Equilibrium. — Evidently, in dealing with 
a system of concurrent forces, it would be a simple matter to 



10 MECHANICS OF ENGINEERING. 

replace any two of the forces by their resultant (diagonal 
formed on them), then to combine this resultant with a third 
force, and so on until all the forces had been combined, the 
last resultant being the resultant of the whole system. The 
foregoing treatment, however, is useful in showing that for 
equilibrium of concurrent forces in a plane only two conditions 
are necessary, viz., 2X = and 2 Y = 0; while in space 
there are three, 2X = 7 27= 0, and 2Z = 0. In Case I., 
then, we have conditions enough for determining two unknown, 
quantities ; in Case II., three. 

18. Problems involving equilibrium of concurrent forces. 
(A rigid body in equilibrium under no more than three forces 
may be treated as a material point, since the (two or) three 
forces are necessarily concurrent.) 

Problem 1. — A body weighing G lbs. rests on a horizontal 

table: required the pressure between it and the table. Fig. 8. 

Consider the body free, i.e., conceive all other bodies removed 

, (the table in this instance), being replaced by the 

forces which they exert on the first body. Taking 



i 



the axis Y vertical and positive upward, and not 
--+X assuming in advance either the amount or direc- 
ts tion of iT, the pressure of the table against the 
body, but knowing that G, the action of the earth 
fig. 8. on ^ ie body, is vertical and downward, we have 
here a system of concurrent forces in equilibrium, in which 
the X and Y components of G are known (being and — 
G respectively), while those, iT x and iT T , of N are unknown. 
Putting 2X — 0, we have ^ + 0=0; i.e., iThas no hori- 
zontal component, .-. JSF is vertical. Putting 2 Y = 0, we 
have JV y — G = 0, .*. JV Y = -f- G ; or the vertical component 
of iV, i.e., JV itself, is positive (upward in this case), and is 
numerically equal to G. 

Peoblem 2. — Fig. 9. A body of weight G (lbs.) is moving 
in a straight line over a rough horizontal table w T ith a uniform 
velocity c (feet per second) to the right. The tension in an 
oblique cord by which it is pulled is given, and = P (lbs.),. 



STATICS OF A MATERIAL POINT. 11 

which remains constant, the cord making a given angle of 
elevation, or, with the path of the body. Required the vertical 
pressure JV (lbs.) of the table, and also its + y 

horizontal action F (frietion) (lbs.) against p 

the body. ^ Jl_j^y 

Referring by anticipation to Newton's first ~^j""^"~J" " 
law of motion, viz., a material point acted w^nwfljfcr 
on by no force or by balanced forces is either FlG - 9 - 

at rest or moving uniformly in a straight line, we see that this 
problem is a case of balanced forces, i.e., of equilibrium. Since 
there are only two unknown quantities, JV and F, we may 
determine them by the two equations of Case I., taking the 
axes X and Y as before. Here let us leave the direction of 
iTas well as its amount to be determined by the analysis. As 
i^must evidently point toward the left, treat it as negative in 
summing the X components ; the analysis, therefore, can be 
expected to give only its numerical value. 
2X = gives P cos a — F— 0. .*. F = P cos a. 

2Y= gives JV+P sin a- G = 0. .'. JV '= G - Psin a. 
.'. JV is upward or downward according as G is > or < P 
sin a. For JV to be a downward pressure upon the bod} 7 would, 
require the surface of the table to be above it. The ratio of the 
friction F to the pressure JV which produces it can now be 
obtained, and is called the coefficient of friction. It may vary 
slightly with the velocity. 

This problem may be looked upon as arising from an experi- 
ment made to determine the coefficient of friction between the 
given surfaces at the given uniform velocity. 

19. The Free-Body Method. — The foregoing rather labored so- 
lutions of very simple problems have been made such to illus- 
trate what may be called the free-body method of treating any 
problem involving a body acted on by a system of forces. It 
consists in conceiving the body isolated from all others which 
act on it in any way, those actions being introduced as so many 
forces, known or unknown, in amount and position. The sys- 
tem of forces thus formed may be made to yield certain equa- 



12 MECHANICS OF ENGINEERING. 

tions, whose character and number depend on circumstances, 
such as the behavior of the body, whether the forces are con- 
fined to a plane or not, etc., and which are therefore theoreti- 
cally available for determining an equal number of unknown 
quantities, whether these be forces, masses, spaces, times, or 
abstract numbers. Of course in some instances the unknown 
quantities may enter these equations with such high powers 
that the elimination may be impossible ; but this is a matter 
of algebra, not of mechanics. 



PARALLEL FORCES AND THE CENTRE OF GRAVITY. 13 



CHAPTER II. 

PARALLEL FORCES AND THE CENTRE OF GRAVITY. 

20. Preliminary Remarks. — Although by its title this section 
should be restricted to a treatment of the equilibrium of forces, 
certain propositions involving the composition and resolution 
of forces, without reference to the behavior of the body under 
their action, will be found necessary as preliminary to the prin- 
cipal object in view. 

As a rigid body possesses extension in three dimensions, to 
deal with a system of forces acting on it we require three co- 
ordinate axes : in other words, the system consists of " forces 
in space," and in general the forces are non-concurrent. In 
most problems in statics, however, the forces acting are in one- 
plane: we accordingly begin by considering non-concurrent 
forces in a plane, of which the simplest case is that of two* 
parallel forces. For the present the body on which the forces 
act will not be shown in the figure, but must be understood to* 
be there (since we have no conception of forces independently 
of material bodies). The device will frequently be adopted of 
introducing into the given system two opposite and equal forces 
acting in the same line : evidently this will not alter the effect 
of the given system, as regards the rest or motion of the body. 

P'!\ — F 

21. Resultant of two Parallel ; \ 



Forces. j 

Case I. — The two forces have £— J~- A 

,. . -r^. ,,„ O S A\ ;D /B S 

the same direction. Jbig. 10. 



Q Q 



« \-a. -y "7[ 

~i I L...j 



Wr\ 



Let P and Q be the given forces, 

and AB a line perpendicular to I \= 

them (P and Q are supposed to have si*--- ~^*-- »l s 

been transferred to the intersections FlG - 10 - 

A and B). Put in at A and B two equal and opposite- 
forces S and S, combining them with P and Q to form P r 



14 MECHANICS OF ENGINEERING. 

and Q '. Transfer P' and Q' to their intersection at C, and there 
resolve them again into S and P, S and Q. S and S annul each 
other at C\ therefore P and Q, acting along a common line CD, 
replace the P and Q first given ; i.e., the resultant of the origi- 
nal two forces is a force P =P -f- Q, acting parallel to them 
through the point D, whose position must now be determined. 
The triangle CAD is similar to the triangle shaded by lines, 
.*. P : 8 :: CD : x; and CDB being similar to the triangle 
shaded by dots, .*. S : Q :: a — x : CD. Combining these, we 

P a — x -, Qa Qa 

have 7^ = an(1 •*• « = -^rr7% ~ ^> • Now write tnis 

Px = Qa, and add Pc, i.e., J 3 ^ -f- Q c -> to eacn member, <? being 
the distance of (Fig. 10), any point in AB produced, from 
A. This will give P(x -f- c) = Pc -\-Q(a -\- c), in which e, 
# + c, and x -\- c are respectively the lengths of perpendiculars 
let fall from upon P, Q, and their resultant P. Any one of 
these products, such &sPc, is for convenience (since products of 
this form occur so frequently in Mechanics as a result of alge- 
braic transformation) called the Moment of the force about the 
arbitrary point 0. Hence the resultant of two parallel forces of 
the same direction is equal to their sum, acts in their plane, in 
a line parallel to them, and at such a distance from any arbi- 
trary point in their plane as may be determined by writing 
its moment about equal to the sum of the moments of the 
two forces about 0. is called a centre of moments, and each 
of the perpendiculars a lever-arm. 

Case II. — Two parallel forces P and Q of opposite direc- 
tions. Fig. 11. By a process similar to the foregoing, we 

obtain P = P - Q and \p - Q)x 

= Qa, i.e., Px = Qa. Subtract 

each member of the last equation 

from Pc (i.e., Pc— Qc), in which c 

\ is the distance, from A, of any arbi- 

^v trary point O in AB produced. This 

T_~ gi ves J£(q _ r^ — p c _ Q^ a _|_ c y 

~% a v ™\ But (c — x), c, and (a -f- c) are re- 

fig. n. spectively the perpendiculars, from 



PARALLEL FORCES AND THE CENTRE OF GRAVITY. 15 

O, upon P, P, and Q. That is, P(c — a?) is the moment of R 
about 0; Pc, that of P about 0\ and $(<3+c), tnat °f C 
about (?. But the moment of Q is subtracted from that of P, 
which corresponds with the fact that Q in this figure would 
produce a rotation about opposite in direction to that of P. 
Having in view, then, this imaginary rotation, we may define 
the moment of a force as positive when the indicated direction 
about the given point is against the hands of a watch; as nega- 
tive when with the hands of a watch. 

Hence, in general, the resultant of- any two parallel forces is, 
in amount, equal to their algebraic sum, acts in a parallel direc- 
tion in the same plane, while its moment, about any arbitrary 
point in the plane, is equal to the algebraic sum of the mo- 
ments of the two forces about the same point. 

Corollary. — If each term in the preceding moment equations 
be multiplied by the secant of an angle (<*, Fig. 12) thus : 





Fig. 12. 



Fig. 13. 



(using the notation of Fig. 12), we have Pa sec a = P x a x 
sec a -f- P 2 a 2 sec <x, i.e., Pb = P 1 b 1 -f- P 2 & 2 , in which b, b 19 
and b 2 are the oblique distances of the three lines of action 
from any point in their plane, and lie on the same straight 
line ; P is the resultant of the parallel forces P x and P a . 



22. Resultant of any System of Parallel Forces in Space. — 
Let P l9 P 2 , P 3 , etc., be the forces of the system, and x t , y„ 
s„ a?„ y„ s a , etc., the co-ordinates of their points of application 
as referred to an arbitrary set of three co-ordinate axes X, l 7 ^ 
and Z, perpendicular to each other. Each force is here re- 



16 MECHANICS OF ENGINEERING. 

stricted to a definite point of application in its line of action 
(with reference to establishing more directly the fundamental 
equations for the co-ordinates of the centre of gravity of a 
body). The resultant P' of any two of the forces, as 
P l and P v is ±= P l -\- P a , and may be applied at (7, the in- 
tersection of its own line of action with a line BD joining 
the points of application of P x and P a , its components. 
Produce the latter line to A, where it pierces the plane XY, 
and let b„ b\ and & 2 , respectively, be the distances of B, C, 
P, from A ; then from the corollary of the last article we have 

Pb' = PA + PA; 

but from similar triangles 

V : l x : \ :: z' : z x : s„ .-. P'z' = P x z x + P 2 z 2 . 

Now combine P', applied at (7, with P 8 , applied at E, calling- 
their resultant P" and its vertical co-ordinate z'\ and we obtain 

P"z" = P'z' + P z z„ i.e., P"z" = P x z x + P A + P 3 2 3 , 
also 

P" = p' + p = p 2 + p 2 + p s . 

Proceeding thus until all the forces have been considered, we 
shall have finally, for the resultant of the whole system, 

B = P, + P,+ P s + etc; 

and for the vertical co-ordinate of its point of application, 

which we may write z, 

Pz = P x z x + P A + P,s, + etc ; 

P x z x + P,s, + P % z % . . . _ 2(Pz) 

P. + P.+ P.+ .-.. ~ 2P 5 

and similarly for the other co-ordinates. 

- 2(Px) J - 2(Py) 
» = 2j p and y = -^p-. 

In these equations, in the general case, such products as P x z xy 
etc., cannot strictly be called moments. The point whose co- 



PARALLEL FORCES AND THE CENTRE OE GRAVITY. 17 

ordinates are the x, y, and 2, just obtained, is called the Centre 
of Parallel Forces, and its position is independent of the (com- 
mon). direction, of the forces concerned. 

Example. — If the parallel forces are contained in one plane, 
and the axis Ybe assumed parallel to the direction of the 
forces, then each product like P x x x will be a moment, as de- 
fined in § 21 ; and it will be noticed in the accompanying nu- 
merical example, Fig. 14, that a detailed substitution in the 
equation P 3 ^ |Y R ^ 

Rx = P^ + P^+etc, . . .(1) U...l,,j"H 3 

having regard to the proper sign of each , , Oj +X 

force and of each abscissa, gives the same Fig. 14. 

result as if each product Px were first obtained numerically, 
and a sign affixed to the product considered as a moment 
about the point 0. Let P 1 = — 1 lb.; P 2 = + 2 lbs.; P s = 
+ 3 lbs.; P 4 = -6 lbs.; x x = + 1 ft.; x, = + 3 ft.; a>, = - 2 
ft.; and a? 4 = — 1 ft. Required the amount and position of the 
resultant R. In amount R = 2P = — l_|_2-f-3 — 6 = — 2 
lbs.; i.e., it is a downward force of 2 lbs. As to its position, 
Rx= 2(Px) gives ( - 2)5 = ( — 1) X ( + 1) + 2 X 3 + 
3 X (- 2) + (—6) X(— l) = —l + 6 — 6 + 6. Now from 
the figure, by inspection, it is evident that the moment of P l 
about O is negative (with the hands of a watch), and is numer- 
ically = 1, i.e., its moment = — 1 ; similarly, by inspection, 
that of P 2 is seen to be positive, that of P 3 negative, that of 
P 4 positive; which agree with the results just found, that 
(- 2)x = — 1 + 6 — 6 + 6 = + 5 ft. lbs. (Since a moment 
is a product of a force (lbs.) by a length (ft.), it may be called 
so many foot-pounds.) Next, solving for x, we obtain 
x = (+ 5) -4- .(— 2) = — 2.5 ft.; i.e., the resultant of the given 
forces is a downward force of 2 lbs. acting in a vertical line 
2.5 ft. to the left of the origin. Hence, if the body in question 
be a horizontal rod whose weight has been already included in 
the statement of forces, a support placed 2.5 ft. to the left of 
and capable of resisting at least 2 lbs. downward pressure 
will preserve equilibrium ; and the pressure which it exerts 



t'H r 



18 MECHANICS OF ENGINEERING. 

against the rod must be an upward force, P 6 , of 2 lbs., i.e. ^e 
equal and opposite of the resultant of P„ jP 2 , P s , P A . 

Fig. 15 shows the rod as a free body in equilibrium -trader 
the five forces. P & = -\- 2 lbs. = the reaction of the support. 

Of course P b is one of a pair of equal 
and opposite forces ; the other one 
— j is the pressure of the rod against the 
P 5 [ — 2.5— -i<D support, and would take its place among 

fig. 15. the f orces acting on the support. 

23. Centre of Gravity. — Among the forces acting on any 
rigid body at the surface of the earth is the so-called attraction 
of the latter (i.e., gravitation), as shown by a spring-balance, 
which indicates the weight of the body hung upon it. The 
weights of the different particles of any rigid body constitute a 
system of parallel forces (practically so, though actually slightly 
convergent). The point of application of the resultant of these 
forces is called the centre of gravity of the body, and may also 
be considered the centre of m.ass, the body being of very small 
dimensions compared with the earth's radius. 

If x, y, and z denote the co-ordinates of the centre of gravity 
of a body referred to three co-ordinate axes, the equations 
derived for them in § 22 are directly applicable, with slight 
changes in notation. 

Denote the weight of any particle of the body by dG, its 
volume by dV, by y its heaviness (rate of weight, see § 7) and 
its co-ordinates by x, y, and z ; then, using the integral sign as 
indicating a summation of like terms for all the particles of the 
body, we have, for heterogeneous bodies, 

fyxd V m 7t _ fyyd V m - _ fyzd V^ 
fydV' y fydV 9 fydV * * w 

while, if the body is homogeneous, y is the same for all its ele- 
ments, and being therefore placed outside the sign of summa- 
tion, is cancelled out, leaving for homogeneous bodies ( V de- 
noting the total volume) 

- fxdV - fydV - fzdV 



PARALLEL FORCES AND THE CENTRE OF GRAVITY. 19 

Corollary. — It is also evident that if a homogeneous body is 
for convenience considered as made up of several finite parts, 
whose volumes are V„ V 2 , etc., and whose gravity co-ordinates 
are a? 13 y v z x ; a? a , y 3 , s 2 ; etc., we may write 



Vp x + V,x 2 + 



x = 



F + F + 



(3) 



If the body is heterogeneous, put G x (weights), etc., instead 
of V„ etc., in equation (3). 

If the body is an infinitely thin homogeneous shell of uni- 
form thickness = h 4 then d V = hdF(dF denoting an element, 
and Fihe whole area of one surface) and equations (2) become, 
after cancellation, 

x — p 9 y — f 9 z — F • • • • W 

Similarly, for a homogeneous wire of constant small cross- 
section (i.e.. a geometrical line, having weight), its length 
being s, and an element of length ds, we obtain 

X ~ S ' V ~ 8 > Z ~ S ' ' ' ' W 

It is often convenient to find the centre of gravity of a thin 
plate by experiment, balancing it on a needle-point ; other 
shapes are not so easily dealt with. 

24. Symmetry. — Considerations of symmetry of form often 
determine the centre of gravity of homogeneous solids without 
analysis, or limit it to a certain line or plane. Thus the centre 
of gravity of a sphere, or any regular polyedron, is at its centre 
of figure ; of a right cylinder, in the middle of its axis ; of a 
thin plate of the form of a circle or regular polygon, in the 
centre of figure ; of a straight wire of uniform cross-section, in 
the middle of its length. 

Again, if a homogeneous body is symmetrical about a plane, 
the centre of gravity must lie in that plane, called a plane of 



20 MECHANICS OF ENGINEERING. 

gravity; if about a line, in that line called a line of gravity;: 
if about a point, in that point. 

25. By considering certain modes of subdivision of a homo- 
geneous body, lines or planes of gravity are often made appar- 
ent. E.g., a line joining the middle of the bases of a trape- 
zoidal plate is a line of gravity, since it bisects all the strips 
of uniform width determined by drawing parallels to the 
bases ; similarly, a line joining the apex of a triangular plate to- 
the middle of the opposite side is a line of gravity. Other 
cases can easily be suggested by the student. 

26. Problems. — (1) Kequired the position of the centre of 

A gravity of 2^ fine homogeneous wire of the 

| Y / j\-T3 V form of a circular arc, AB, Fig. 16. Take 

y- y l-"'C<j.\\ the origin at the centre of the circle, and 

U^'''""" ill the axis X bisecting the wire. Let the 

u X. dxi j length of the wire, s, = 2s x ; ds = ele- 

\. ~f I «'i ment of arc. We need determine only the 

§^/ x, since evidently y — 0. Equations (5) r 

/hjds 

fig. 16. g 23, are applicable here, i.e., x = . 

From similar triangles we have 

rdy 
ds :dy :: r : x; .*. ds = — -; 

x 

:.~x — Z- Pdv — -x—n i.e., = chord X radius -f- length of 

wire. For a semicircular wire, this reduces to x = 2r -f- n. 

Problem 2. Centre of gravity of trapezoidal (and trian- 
gular) thin plates, homogeneous, etc. — Prolong the non-parallel 
sides of the trapezoid to intersect at 0, which take as an origin,, 
making the axis X perpendicular to the bases b and l x . ~We 
may here use equations (4), § 23, and may take a vertical strip 
for our element of area, dF, in determining x ; for each point 
of such a strip has the same x. Now dF = (y -f- y')dx, and 



PARALLEL FORCES AND THE CENTRE OP GRAVITY. 21 



from similar triangles y -f- y' = j x. Hence F = <r (bh — bji^) 



can be written — 7 (A 2 



2A 
"b Ph 



A/), and x = - ^ becomes 



x*dx -f 



lb 

2 A 



(A 2 - h?) = 



3 A 2 - A, 



for the trapezoid. 



For a triangle A x = 0, and we Lave x— A ; that is, the 

o 

centre of gravity of a triangle is one third the altitude from the 

base. The centre "of gravity is finally determined by knowing 



• — * 






•x 






y 


b 


( ^^-----^- 


-X-i 


y' 


1 

i 




Fig. 17. 



Fig. 18. 



that a line joining the middles of b and b l is a line of gravity; 
or joining O and the middle of b in the case of a triangle. 

Problem 3. Sector of a circle. Thin plate, etc. — Let the 
notation, axes, etc., be as in Fig. 18. Angle of sector = 2a; 
x = ? Using polar co-ordinates, the element of area dF (a 
small rectangle) = pdcp . dp, and its x = p cos cp ; hence the 
total area = 

i.e., F = r 2 a. From equations (4), § 23, we have 
x = yfxdF 

— yJ J cos <PP*dpdp — j?J_ a [_ COS( pJ P'dpj dcp. 



22 MECHANICS OF ENGINEERING. 

(Note on double integration, — The quantity 
cos cp J p* dp \d<p, 

is that portion of the summation / / cos cpp'dpdcp which 

belongs to a single elementary sector (triangle), since all its- 
elements (rectangles), from centre to circumference, have the 
same cp and dcp.) 
That is, 



1 r 3 p + a r * r+a 2 r sin a 

^ ■ 3 «/- - C0B ^ = Wa L™ * = 3 • ~^T ' 



- 1 r 

x 



a i i i. ~ ^ r Sln i P 

or, putting p = 2a = total angle oi sector, a? = „■ 3 — ► 

6 p 

— 4^ 

For a semicircular plate this reduces to x = o - . 

[iVtffe. — In numerical substitution the arcs or and /? used 
above (unless sin or cos is prefixed) are understood to be ex- 
pressed in circular measure (7r-measure) ; e.g., for a quad- 
rant, fi = | = 1.5707 + ; for 30°,/? = |; or, in general, if p 

. -, 180 ° 1 *~\ 

in degrees = , then p in ^--measure = -. 

\ Problem 4. Sector of a flat ring ; thin 

c \J a plate, etc. — Treatment similar to that of 

Problem 3, the difference being that the 




Fio. 19. 



J limits of the interior integrations are 
instead of . Result, 

l— n 



— 4 r* — r* sin \fi 
X = 3 ' r? - r; ' ~~~W~ % 



Parallel forces and the centre of gravity. 23 
Pboblem 5. — Segment of a circle; thin plate, etc. — Fig. 20. 



Since each rectangular element of any ver- 
tical strip has the same x, we may take the 
strip as dF in finding x, and use y as the 
half -height of the strip. dF = 2ydx, and 
from similar triangles x : y : : ( — dy) : dx, 
i.e., xdx = — ydy. Hence from eq. (4), 
§23, 



- fxdF 

x --p- 



fxZydx 
~~~F~ 



VM 



F 



2_ 
ZF 



} &.x 




but a = the half-chord, hence, finally, x = -tajp " • 

Peoblem 6. — Trapezoid; thin plate, etc., 
by the method in the corollary of § 23 ; equa- 
tions (3). Kequired the distance x from the 
base AB. Join DB, thus dividing the trape- 
zoid ABCD into two triangles ADB = F x 
and DBG = F„ whose gravity x J s are, re- 
spectively, x x = -JA and a? 2 = f h. Also, F x 
= ihb x , F 7 = -JA5 2 , and F (area of trape- 
zoid) = ±h(b x + 5 2 ). Eq. (3) of § 23 gives 
Fxj= F x x x -f- i^ 2 a? 2 ; hence, substituting,^ -j- 
b,)x=ib x h+%b 2 h. 




Fig. 21. 



-_h (ft, + 26,) 

• w — « • t ; t « 



The line joining the middles of b x and b 2 is a line of gravity, and 
is divided in such a ratio by the centre of gravity that the fol- 
lowing construction for finding the latter holds good : Prolong 
each base, in opposite directions, an amount equal to the other 
base; join the two points thus found: the intersection with 
the other line of gravity is the centre of gravity of the trape- 
zoid. Thus, Fig. 21, with BF= b a and DF= b x , join FF, 
etc. 



24 MECHANICS OF ENGINEERING. 

Problem 7. Homogeneous oblique cone or pyramid. — 
Take the origin at the vertex, and the axis X perpendicular to 
the base (or bases, if a frustum). In finding x we may put 
d V = volume of any lamina parallel to YZ, F being the base 
of such a lamina, each point of the lamina having the same x. 
Hence, (equations (2), § 23), 



but 



and 



= \fxdV, V=fdV=fFdx; 



F:F % ::«?:h;., .:F=fix', 



h. 



q ~L 4 7 4 

For a frustum, x — — . ~- % ~ ; while for a pyramid, h lt be- 

— 3 
ing = 0, x = -rh. Hence the centre of gravity of a pyramid 

is one fourth the altitude from the base. It also lies in the line 

- .ft™ --^K joining the vertex to the centre of gravity 

_/[ d 1 of the base. 

yv-J^T Problem 8. — If the heaviness of the ma- 

^>>^\ x ~a Serial °f tne above cone or pyramid varied 

~/ y directly as x, y^ being its heaviness at the 

fig. 22. Dase F^ we would use equations (1), § 23, 

putting y = j 3 x ; and finally, for the frustum, 



_ 4 h;-h* 

x~ 



5 • h: - v 



4 

and for a complete cone x = —h r 



27. The Centrobaric Method. — If an elementary area dF be 
revolved about an axis in its plane, through an angle a < 2tt. 



PARALLEL FORCES AND THE CENTRE OF GRAVITY. 25 




the distance from the axis being = x, the volume generated is 
d V = axdF, and the total volume generated by all the dF's 
of a finite plane figure whose plane con- 
tains the axis and which lies entirely on one 
side of the axis, will be V = fdV = 
afxdF. But from §23, afxdF^ aFx; 
ax being the length of path described by 
the centre of gravity of the plane figure, 
we may write : The volume of a solid of revolution generated 
by a plane figure, lying on one side of the axis, equals the 
area of the figure multiplied by the length of curve described 
by the centre of gravity of the figure. 

A corresponding statement may be made for the surface 
generated by the revolution of a line. The arc a must be ex- 
pressed in n measure in numerical work. 



27a. Centre of 
.B 




Fig. 23a. 



Gravity of any Quadrilateral. — Fig. 23$. 
Construction; ABGB being any quad- 
rilateral. Draw the diagonals. On the 
long segment BK of BB lay off BE = 
BK, the shorter, to determine E; simi- 
larly, determine N on the other diagonal, 
by making GN = AK. Bisect EK in H 
and KN'm M. The intersection of EM 
and NH is the centre of gravity, C. 
Proof— H being the middle of BB, and AH and HG 
having been joined, I the centre of gravity of the triangle 
ABB is found on AH, by making HI — \AH; similarly, by 
making HL — \HG, L is the centre of gravity of triangle 
BBG. . * . IL is parallel to AG and is a gravity-line of the 
whole figure ; and the centre of gravity C may be found on it 
if we can make CL : CI :: area ABB : area BBG (§ 21). 
But since these triangles have a common base BB, their areas 
are proportional to the slant heights (equally inclined to BB) 
AK and KG, i.e., to GN and NA. Hence HN, which di- 
vides IL in the required ratio, contains C, and is .*. a gravity- 
line. By similar reasoning, using the other diagonal, AG, and 



26 MECHANICS OF ENGINEEKING. 

the two triangles into which it divides the whole figure, we- 
may prove EM to be a gravity- line also. Hence the construc- 
tion is proved. 

27b. Examples. — 1. Required the volume of a sphere by 
the centrobaric method. 

A sphere may be generated by a semicircle revolving about 
its diameter through an arc a = 2?r. The length of the path 

described by its centre of gravity is = 2tt^— (see Prob. 3, § 

26), while the area of the semicircle is ^jtt\ Hence by § 27,, 

4:7* 4 

Yolume generated = 27f . — . %7ir* = « W. 

2. Required the position of the centre of gravity of the sector 
of a flat ring in which r x = 21 feet, r a = 20 feet, and /3 = 80° 
(see Fig. 19, and § 26, Prob. 4). 

sin — = sin 40° = 0.64279, and J3 in circular measure = 

— - n — — 7t = 1.3962. By using r x and r a in feet, x will be 
obtained in feet. 



_ 4 r' — rf Sin 2 4 12610.64279 



% '.X = 7T • 



3 ' r, 2 - r*' ft ~ 3 * 41 * 1.3962 



~- = 18.87 feet. 



STATICS OF A RIGID BODY. 27 



CHAPTER III. 

STATICS OF A RIGID BODY. 

28. Couples. — On account of the peculiar properties and 
utility of a system of two equal forces acting in parallel lines 
and in opposite directions, it is specially /t? 
considered, and called a Couple. The /^ f 
arm of a couple is the perpendicular fcs^ -'f^L 
distance between the forces ; its moment, \ -4a Joj^; J^p 
the product of this arm, by one of the <^j ^^^ ^<^^ 
forces. The axis of a couple is an ^^^^^^^ 
imaginary line drawn perpendicular to Fia . 2 4. 

its plane on that side from which the rotation appears positive 
(against the hands of a watch). (An ideal rotation is meant, 
suggested by the position of the arrows ; any actual rotation 
of the rigid body is a subject for future consideration.) In 
dealing with two or more couples the lengths of their axes are 
made proportional to their moments; in fact, by selecting a 
proper scale, numerically equal to these moments. E.g., in Fig. 
24, the moments of the two couples there shown are Pa and 
Qb\ their axes p and q so laid off that Pa : Qb :: p : q, and 
that the ideal rotation may appear positive, viewed from the 
outer end of the axis. 

29. No single force can balance a couple. — For suppose the 
couple P, P, could be balanced by a force R' , then this, acting 

p* at some point C, ought to hold the couple 

dL._ /:.:^_ B C__ in equilibrium. Draw CO through 0, the 

*T /p j^t centre of symmetry of the couple, and 

fig. 25. make OP =OC. At D put in two op- 

posite and equal forces, S and T, equal and parallel to i?'. 
The supposed equilibrium is undisturbed. But if R\ P, and 



28 MECHANICS OF ENGINEERING. 

P are in equilibrium, so ought (by symmetry about 0) #, P, 
and P to be in equilibrium, and they may be removed without 
disturbing equilibrium. But we have left Tand'i?', which are 
■evidently not in equilibrium ; .*. the proposition is proved by 
this reductio ad absurdum. Conversely a couple has no single 
resultant. 

30. A couple may he transferred anywhere in its own plane. 
— First, it may be turned through any angle a, about any 

pi point of its arm, or of its arm produced. 

G T~-'i :'a ■["" Let (P, P')he a couple, G any point of its 

\.- ; 'SJ?-i fp' arm (produced), and a any angle. Make 

^U : ^ x GC = GA, CD = AB, and put in at C, 

P \ C \ 2 *V I P a and P 2 equal to P (or P'), opposite to 

q\ Pt - : -k"f each other and perpendicular to GO; and 

3 \ r* P z and P A similarly at D. Now apply and 

fig. 26. combine P and P x at 0, P' and P< at 0'; 

then evidently 11 and R r neutralize each other, leaving P % and 
P 2 equivalent to the original couple (P, P'). The arm 
CD = AB. Secondly, if G be at infinity, and a = 0, the 
same proof applies, i.e., a couple may be moved parallel to 
itself in its own plane. Therefore, by a combination of the 
-two transferrals, the proposition is established for any trans- 
ferral in the plane. 

31. A couple may he replaced hy another of equal moment 
■in a parallel plane. — Let (P, P') be a couple. Let CD, in a 
parallel plane, be parallel to AB. At D put in a pair of equal 

AP 

and opposite forces, S s and S A , parallel to P and each = =P. 

ED 

BE 
Similarly at C, $ and S a parallel to P and each = ==P. 

JBut, from similar triangles, 

AE — BE. . Q — Q — Q — Q 
j^jy — PC* ' * * — 2 — 3 — 4 * 



STATICS OF A RIGID BODY. 29 

[Note. — The above values are so chosen that the intersection point E 
may be the point of application of (P' -f- S 2 ), the resultant of P and & 9 ;.. 
and also of (P-{-S 3 ), the resultant of Pand S a , as follows from § 21; thus 
(Fig. 28), R, the resultant of the two parallel forces Pand # 3 , is = P-\-S z , 
and its moment about any centre of moments, as E, its own point of ap- 
plication, should equal the (algebraic) sum of the moments of its com- 

AE 

ponents about E; i.e., R X zero =P.AE — S 3 . JJE; .:S a = = . P.] 



DE 



IA 



cr 



I i 

si-., 



p" 



...I s *. 1 T 

D E A 



(P-S 2 ) 
Fig. 27. Fig. 28. 

Keplacing P' and tf, by (P' + #,), and P and tf, by- 
(-P -f- S 9 ), the latter resultants cancel each other at E, leavings 
the couple (S v SJ with an arm CD, equivalent to the original 

couple P, P' with an arm AB. But, since S 1 = == . P — 

AlB 

== . P, we have S.XCD = PxAB ; that is, their moments 

C/JJ 

are equal. 

32. Transferral and Transformation of Couples.— In view of 
the foregoing, we may state, in general, that a couple acting on. 
a rigid body may be transferred to any position in any parallel 
plane, and may have the values of its forces and arm changed 
in any way so long as its moment is kept unchanged, and still 
have the same effect on the rigid body (as to rest or motion, 
not in distorting it). 

Corollai'ies. — A couple may be replaced by another in any 
position so long as their axes are equal and parallel and simi- 
larly situated with respect to their planes. 

A couple can be balanced only by another couple whose axis 
is equal and parallel to that of the first, and dissimilarly situ- 
ated. For example, Fig. 29, Pa being = Qb, the rigid body 
AB (here supposed without weight) is in equilibrium in each 



30 



MECHANICS OF ENGINEERING. 



case shown. By " reduction of a couple to a certain arm #" 
is meant that for the original couple whose arm is a', with 
forces each = P', a new couple is substituted whose arm shall 



be 



and the value of whose forces P and P must be com- 



puted from the condition 
Pa = P'a', 



i.e., 



P = P'a! 




Fig. 29. 



Fig. 30. 



33. Composition of Couples. — Let (P, P f ) and ($, Q f ) be two 

couples in different planes reduced to the same arm AB — a, 
which is a portion of the line of intersection of their planes. 
That is, whatever the original values of the individual forces 
and arms of the two couples were, they have been transferred 
and replaced in accordance with § 32, so that P.AB, the 
moment of the first couple, and the direction of its axis, p, 
have remained unchanged ; similarly for the other couple. 
Combining P with Q and P' with Q\ we have a resultant 
couple (P, P') whose arm is also AB. The axes p and q of 
the component couples are proportional to P . AB and Q . AB, 
i.e., to P and Q, and contain the same angle as P and Q. 
Therefore the parallelogram p . . . q is similar to the parallelo- 
gram P . . . Q ; whence p : q : r::P : Q : i?, or p : q : r : : 
Pa : Qa : Pa. Also r is evidently perpendicular to the plane 
of the resultant couple (i?, B ; ), whose moment is Pa. Hence 
r, the diagonal of the parallelogram on p and q, is the axis of 
the resultant couple. To combine two couples, therefore, we 
have only to combine their axes, as if they were forces, by a 
parallelogram, the diagonal being the axis of the resultant 
couple ; the plane of this couple will be perpendicular to the 



STATICS OF A RIGID BODY. 



31 



axis just found, and its moment bears the same relation to the 
moments of the component couples as the diagonal axis to the 
two component axes. Thus, if two couples, of moments Pa 
and Qb, lie in planes perpendicular to each other, their result- 
ant couple has a moment Re — \/(P a f + {Qbf' 

If three couples in different planes are to be combined, the 
axis of their resultant couple is the diagonal of the parallelo- 
piped formed on the axes, laid off to the same scale and point- 
ing in the proper directions, the proper direction of an axis 
being away from the plane of its couple, on the side from 
which the couple appears of positive direction. 

34. If several couples lie in the same plane their axes are 
parallel and the axis of the resultant couple is their algebraic 
sum ; and a similar relation holds for the moments : thus, in 
Fig. 24, the resultant of the two couples has a moment = Qb 
— Pa, which shows us that a convenient way of combining 
couples, when all in one plane, is to call the moments positive 
or negative, according as the ideal rotations are against, or with, 
the hands of a watch, as seen from the same side of the plane ; 
the sign of the algebraic sum will then show the ideal rotation 
of the resultant couple. 

35. Composition of Non-concurrent Forces in a Plane. — Let 

P 15 P 2 , etc., be the forces of the system ; x x , y v x„ y„ etc., the 



Y, 



X,/4 



m 




"Y/ o 

Fig. 31. Fig. 32. 

co-ordinates of their points of application ; and or,, <* 2 , . . . etc., 
their angles with the axis X. Replace P 1 by its components 
J3l, and Y x , parallel to the arbitrary axes of reference. At the 
origin put in two forces, opposite to each other and equal and 
parallel to X x ; similarly for T v (Of course X x = P x cos a and 
Y x = P x sin a.) "We now have P x replaced by two forces X y 



32 MECHANICS OF ENGINEERING. 

and Y x at the origin, and two couples, in the same plane, whose 
moments are respectively — X 1 y l and + T> x , and are there- 
fore (§34) equivalent to a single couple, in the same plane witk 
a moment = (Y 1 x l —X 1 y : ). 

Treating all the remaining forces in the same way, the whole- 
system of forces is replaced by 

the force 2(X) =X X + X, + ... at the origin, along the axis X; 
the force 2( Y) = Y,+ Z" 2 + . . . at the origin, along the axis Y\ 

and the couple whose mom. G = 2 ( Yx — Xy\ which may be 
called the couple G (see Fig. 32), and may be placed anywhere 
in the plane. Now 2(X) and 2( Y) may be combined into a 
force R ; i.e., 

2X 



R = V(^Xf -\- 2 Y) 2 and its direction-cosine is cos a = — p- 

Since, then, the whole system reduces to G and R, we must 

have for equilibrium R = 0, and G = ; i.e., for equilibrium 

2X= 0,2Y=0, and 2(Yx-Xy) = 0. . eq. (1) 

If R alone = 0, the system reduces to a couple whose mo- 
ment is G = 2( Yx—Xy) ; and if G alone = the system re- 
duces to a single force i?, applied at the origin. If, in general, 
neither R nor G = 0, the system is still equivalent to a single 
force, but not applied at the origin (as could hardly be ex- 
pected, since the origin is arbitrary) ; as follows (see Fig. 33) : 

Replace the couple C by one of equal moment, G, with each 

C 1 

force = R. Its arm will therefore be -p. Move this couple 

in the plane so that one of its forces R may cancel the R al- 
ready at the origin, thus leaving a single resultant R for the 
whole system, applied in a line at a perpendicular distance, 

c = -jp , from the origin, and making an angle a whose cosine = 

2X 

-w-j with the axis X. 

36. More convenient form for the equations of equilibrium 
of non-concurrent forces in a plane. — In (I.), Fig. 34, being 



STATICS OF A RIGID BODY. 33 

any point and a its perpendicular distance from a force P ; 
put in at O two equal and opposite forces P and P' = and || 
to P, and we have P replaced by an equal single force P' at 
(9, and a couple whose moment is -f- Pa. (II.) shows a simi- 
lar construction, dealing with the Xsmd ^components of i 3 , 
so that in (II.) P is replaced by single forces X' and Y ! at O 

* /! VL 

/ .<» x ku>x 




Fig. 33. 




(and they are equivalent to a resultant P f , at 0, as in (I.), and 
two couples whose moments are + Yx and — Xy. 

Hence, being the'same point in both cases, the couple Pa 
is equivalent to the two last mentioned, and, their axes being 
parallel, we must have Pa = Yx — Xy. Equations (1), 
§ 35, for equilibrium, may now be written 

2X- 0, 2Y = 0, and 2{Pa) = 0. . . (2) 

In problems involving the equilibrium of non-concurrent 
forces in a plane, we have three independent conditions, or 
equations, and can determine at most three unknown quantities. 
For practical solution, then, the rigid body having been made 
free (by conceiving the actions of all other bodies as repre- 
sented by forces), and being in equilibrium (which it must be 
if at rest), we apply equations (2) literally ; i.e., assuming an 
origin and two axes, equate the sum of the ^components of 
all the forces to zero; similarly for the ^components; and 
then for the " moment-equation," having dropped a perpen- 
dicular from the origin upon each force, write the algebraic 
sum of the products {moments) obtained by multiplying each 
force by its perpendicular, or " lever-arm" equal to zero, call- 
ing each product -f- or — according as the ideal rotation ap- 
pears against, or with, the hands of a watch, as seen from the 
same side of the plane. (The converse convention would do as 
well.) 



34 MECHANICS OF ENGINEERING. 

Sometimes it is convenient to use three moment equations, 
taking a new origin each time, and then the 2X = and ^ Y 
-= are superfluous, as they would not be independent equa- 
tions. 

37. Problems involving Non-concurrent Forces in a Plane.— 

Remarks. The weight of a rigid body is a vertical force 
through its centre of gravity, downwards. 

If the surface of contact of two bodies is smooth the action 
(pressure, or force) of one on the other is perpendicular to the 
surface at the point of contact. If a cord must be imagined 
cut, to make a body free, its tension must be inserted in the 
line of the cord, and in such a direction as to keep taut the 
small portion still fastened to the body. In case the pin of 
a hinge must be removed, to make the body free, its pressure 
against the ring being unknown in direction and amount, it is 
most convenient to represent it by its unknown components X 
and Y, in known directions. In the following problems there 
is supposed to be no friction. If the line of action of an un- 
known force is known, but not its direction (forward or back 
ward), assume a direction for it and adhere to it in all the three 
equations, and if the assumption is correct the value of the 
force, after elimination, will be positive ; if incorrect, negative. 
Problem 1. — Fig. 35. Given an oblique rigid rod, with two 
loads G 1 (its own weight) and (? a ; required the reaction of the 
smooth vertical wall at A, and the direction and amount of the 
Am^-pressure at 0. The reaction at A 
must be horizontal ; call it X. The pres- 
^1 sure at 0, being unknown in direction, will 
have both its X and Y components un- 
known. The three unknowns, then, are 
X , X, and Y , while G t , G» a^ a^ and 
i^ h are known. The figure shows the rod 

Fig. 3o. ag a y ree 00 gy^ a ]i ^he forces acting on it 

have been put in, and, since the rod is at rest, constitute a sys- 
tem of non-concurrent forces in a plane, ready for the condi- 
tions of equilibrium. Taking origin and axes as in the figure, 



6 



STATICS OF A RIGID BODY. 



35 



2X = gives +X - X' = ; 2Y = gives + Y - G, 
— G 2 = ; while 2(Pa) = 0, about 0, gives + X'h — 
G x a x — G 7 a 2 = 0. (The moments of X and Y about 
are, each, = zero.) By elimination we obtain JT = G x -[- 
^ ; JT = X' = [6^ -f- 6r 2 aJ -f- A; while the pressure at 
= VX 2 -\- JT 2 , and makes with the horizontal an angle 
whose tan = Y -i- X . 

[N.B. A special solution for this problem consists in this, that the result- 
ant of the two known forces Gi and Q 2 intersects the line of X in a point 
which is easily found by § 21. The hinge-pressure must pass through this 
point, since three forces in equilibrium must be concurrent.] 

We might vary this problem by limiting X to a safe value, 
depending on the stability of the wall, and making h an un- 
known. The three unknowns would then be X , Y , and h. 

Problem, 2. — Given two rods with loads, three hinges (or 
■" pin -joints"), and all dimensions: required the three hinge- 





Fig. 37. 

pressures; i.e., there are six unknowns, viz., three X and three 
Y components. We obtain three equations from each of the 
two free bodies in Fig. 37. The student may fill out the de- 
tails. Notice the application of the principle of action and 
reaction at B (see § 3). 

Problem 3. — A Warren bridge-truss rests on the horizontal 
smooth abutment-surfaces in Fig. 38. It is composed of equal 
isosceles triangles ; no piece is 
continuous beyond a joint, each 
of which is a join connection. All 
loads are considered as acting at 
the joints, so that each piece will 
be subjected to a simple tension 
or compression. 





f 


p d P 


l« 


0, 


m 


AV 


>u 




*-%°i t 


|E] 


tt 



Fig. 38. 



36 



MECHANICS OF ENGINEERING. 



First, required the reactions of the supports V t and V 2 ; 
these and the loads are called the external forces. 2(Pa), 
about = gives 

V£a - P x .\a- P,.%a- P. . |a = 0; 

while 2(Pa) about K = gives 

- F 1 .3« + P,.i« + P 1 |a + P 1 |a = 0; 

-•- ^ = *[5^ + 3P 2 + P 3 ] ; 
and F 3 = i[P x + 3P 2 + 5PJ. 

Secondly, required the stress (thrust or pull, compression or 
tension) in each of the pieces A, P, and cut by the imaginary 
line PP. The stresses in the pieces are called internal forces. 
These appear in a system of forces acting on a free body only 
when a portion of the truss or frame is conceived separated 
from the remainder in such a way as to expose an internal 
plane of one or more pieces. Consider as a free body the por- 
tion on the left of DE (that on the right would serve as well,, 

but the pulls or thrusts in A, P, and 
C would be found to act in directions- 
opposite to those they have on the 
other portion ; see § 3). Fig. 39. The- 
arrows (forces) A, P, and C are not 
pointed yet. They, with V^ P x , and 
P„, form a system in equilibrium.. 



tt 


offi 


(• 


\fh 


¥X 


1/ \ 


*--Ja — »j 


< -a 



Fig. 39. 



2 (Pa) about O = gives 



(Ah) - VM + P, . *a + P 8 . ia 



0. 



Therefore the moment (Ah) = ia[± V x — 3P, — PJ, which 
is positive, since (from above) 4 V l is > 3P, -f- P a . Hence 
A must point to the left, i.e., is a thrust or compression, and is 

Similarly, taking moments about 0„ the intersection of A 
and P, we have an equation in which the only unknown is O r 
viz., (Oh) - Y$a + P,a = 0. .\ (Oh) = £a[3 V x - 2PJ, 



STATICS OF A RIGID BODY. 



37 



a positive moment, since 3 V x is >2P, ; .*. C must point to the 

right, i.e., is a tension, and = rr-[3 V l — 2PJ. 

Finally, to obtain B, put ^(vert. comps.) = ; i.e. (B cos cp) 
+ V 1 -P 1 -P, = 0. .-. B cos cp = P x + P 2 - V x ; but 
{see foregoing value of JQ we may write 

V x = {P x + P 2 ) - (|P X + *P 3 ) + *i\. 
.-. i? cos 9? will be + (upward) or — (downward), and B will 
be compression or tension, as %P a is < or > \^P l -\-^P^\. 

B = IP + P X -*Q + cos 9 = ^l±^ r l [Pi + Pi _ FJ . 

Problem 4. — Given the weight G x of rod, the weight G„ 

and all the geometrical elements (the student will assume a 

W\(P, 





G, A^ 
Fig. 40. Fig. 41. 

convenient notation); required the tension in the cord, and the 
amount and direction of pressure on hinge-pin. 

Problem 5. — Roof-truss ; pin-connection ; all loads at joints ; 
wind-pressures W and W, normal to OA ; required the three 
reactions or supporting forces (of the two horizontal surfaces 
and one vertical surface), and the 
stress in each piece. All geomet- 
rical elements are given ; also P, 
P P W. 

J. 1? .L 2 , fr . 



38. Composition of Non-concur- 
rent Forces in Space. — Let P x , P„ 
etc., be the given forces, and a? 1? y^ 
z xi x v Vv z v efcc -> their points of ap- 
plication referred to an arbitrary 




Fig. 42. 



origin and axes; or l5 /? 1? y l9 etc., 

the angles made by their lines of application with X, Y, and Z. 



38 



MECHANICS OF ENGINEERING. 



Considering the first force P x , replace it by its three com- 
ponents parallel to the axes, X x = P x cos a x \ Y x = P x cos fi x ; 
and Z x = P x cos y x (P x itself is not shown in the figure). At 
0, and also at A, put a pair of equal and opposite forces, 
each equal and parallel to Z x ; Z x is now replaced by a single 
force Z x acting upward at the origin, and two couples, one 
in a plane parallel to YZ and having a moment = — Z x y x (as 
we see it looking toward from a remote point on the axis 
-f- X), the other in a plane parallel to XZ and having a mo- 
ment = -f- Z x x x (seen from a remote point on the axis -f- Y).. 
Similarly at and C put in pairs of forces equal and parallel 
to X x , and we have X„ at B, replaced by the single force X x 
at the origin, and the couples, one in a plane parallel to XY X 
and having a moment -f- X x y xi seen from a remote point on 
the axis -f- Z, the other in a plane parallel to XZ, and of a 
moment =—X x 3 x , seen from a remote point on the axis -\-Yy 
and finally, by a similar device, Y x at B is replaced by a force 
Y x at the origin and two couples, parallel to the planes XY 
and YZ, and having moments — Y x x x and -f* Y x z^ respective- 
ly. (In Fig. 42 the single forces at the origin are broken 
lines, while the two forces constituting any one of the six 

couples may be recognized as being 
equal and parallel, of opposite di- 
rections, and both continuous, or 
both dotted.) We have, therefore, 
replaced the force P x by three 
forces X x , Y x , Z x , at 0, and six 
couples (shown more clearly in 
Fig. 43; the couples have been 
transferred to symmetrical posi- 
tions). Combining each two couples 
whose axes are parallel to X, Y 9 
or Z, they can be reduced to three, viz., 

one with an X axis and a moment = Y x z x — Z x y x ; 
one with a l^axis and a moment = Z x x x — X x z x ; 
one with a Z axis and a moment = X x y y — Y x x x . 




STATICS OF A RIGID BODY. 39 

Dealing with each of the other forces P„ P % , etc., in the same 
manner, the whole system may finally be replaced by three 
forces 2X, ^Y, and 2Z, at the origin and three couples 
whose moments are, respectively, 

Z = 2( ITz — Zy) with its axis parallel to X\ 
M = 2{Zx — Xz) with its axis parallel to Y; 
2H = ^{Xy — Yx) with its axis parallel to Z. 

The "axes" of these couples, being parallel to the respective 
co-ordinate axes X, Y, and Z, and proportional to the mo- 
ments Z, M, and X, respectively, the axis of their resultant 
C, whose moment "is G, must be the diagonal of a parallelo- 
pipedon constructed on the three component axes (propor- 
tional to) Z, M, and X. Therefore, G = VZ 2 + M 2 + X% 
while the resultant of 2X, 2 Y, and 2Z is 



e = V(zxy + (2 Yy + (szy 

acting at the origin. If <*, /?, and y are the direction-angles 

c-o ^ ^X 2Y 2Z 

of R, we have cos a = -jj~, cos p = ~n~, and cos y = —= ; 

while if A, /*, and v are those of the axis of the couple C, we 

•, L M A & 

have cos A = -p, cos }x = yj, and cos y = -j- y . 

For equilibrium we have both G = and R = 0; i.e., 
separately, six conditions, viz., 

2X-= 0, 2 Y = 0, 2Z=0 ; and Z=0, M=0, X=0 . (1) 

Now, noting that 2X = 0, 2Y= 0, and ^(Xy — Tb) = 
are the conditions for equilibrium of the system of non-concur- 
rent forces which would be formed by projecting each force of 
our actual system upon the plane XY, and similar, relations 
for the planes YZ and XZ, we may restate equations (1) in 
another form, more serviceable in practical problems, viz. : 
Note. — If a system of non-concurrent forces in space is in 
equilibrium, the plane systems formed by projecting the given 
system upon each of three arbitrary co-ordinate planes will each 
be in equilibrium. But we can obtain only six independent 



40 MECHANICS OF ENGINEERING. 

equations in any case, available for six unknowns. If R alone 
= 0, we have the system equivalent to a couple G, whose 
moment = G ; if G alone = 0, the system has a single re- 
sultant R applied at the origin. In general, neither R nor G 
being = 0, we cannot further combine R and C (as was done 
with non-concurrent forces in a plane) to produce a single re- 
sultant unless R and C are in the same plane ; i.e., when the 
angle between R and the axis of (7 is = 90°. Call that angle 
6. If, then, cos 6 = cos a cos X -\- cos (3 cos jn -\- cos y cos v 
is = = cos 90°, we may combine R and G to produce a 
single resultant for the whole system ; acting in a plane con- 
taining R and parallel to the plane of G in a direction parallel 

C 

to R, at a perpendicular distance c = -p from the origin and 

= R in intensity. The condition that a system of forces in 
space have a single resultant is, therefore, substituting the 
previously derived values of the cosines, (2X) . L-\- (2 Y) . M 
+ (2Z) .N=0. 

This includes the cases when R is zero and when the system 
reduces to a couple. 

To return to the general case, R and G not being in the 
same plane, the composition of forces in space cannot be 
further simplified. Still we can give any value we please to 

P, one of the forces of the couple G, calculate the correspond- 
ent 
ing arm a = -p, then transfer G until one of the jP's has the 

same point of application as R, and combine them by the 
parallelogram of forces. We thus have the whole system 
equivalent to two forces, viz., the second P, and the resultant 
of R and the first P. These two forces are not in the same 
plane, and therefore cannot be replaced by a single resultant. 

39. Problem. (Non-concurrent forces in space.) — Given all 
geometrical elements (including <*, /?, y, angles of JP), also the 
weight of Q, and weight of apparatus G ; A being a hinge whose 
pin is in the axis Y, G a ball-and-socket joint : required the 
amount of P (lbs.) to preserve equilibrium, also the pressures 



STATICS OF A RIGID BODY. 



41 



(amount anc* direction) at A and ; no friction. Replace P 
by its X, .F, and Z components. The pressure at A will have 




Fig. 44. 



Z and X components ; that at 0, X, Y y and Z components. 

The body is now free, and there are six unknowns. 
3X, 2 Y, and 2Z give, respectively, 

P cos a + X x + X = ; 

P cos ft + Y = 0; and Z, + Z — Q — G -P cos y = 0. 

As for moment-equations (see note in last paragraph), project- 
ing the system upon YZ and putting 2(Pa) about (9 = 0, 
we have 

- ZJ+ Qd+Ge + (P cos y)b + (Pcos ft)o = ; 
projecting it upon XZ, and putting 2(Pa) about = 0, we 
have Qr — (P cos a)c — (P cos y)a = ; 

projecting on XI 7 ^ moments about (9 give 

X X Z + (P cos a)b - (P cos /?> = 0. 

From these six equations we may obtain the six unknowns, 
T 3 , X , Y , Z , X a , and Z x . If for any one of these a negative 
result is obtained, it shows that its direction in Fig. 44 should 
be reversed. 



42 



MECHANICS OF ENGINEERING. 



CHAPTER IV. 



STATICS OF FLEXIBLE CORDS. 



40. Postulate and Principles. — The cords are perfectly flexi- 
ble and inextensible. All problems will be restricted to one 
plane. Solutions of problems are based on three principles, 
viz.: 

Prin. I. — The strain on a cord at any point can act only 
along the cord, or along the tangent if it be curved. 

Prin. II. — We may apply to flexible cords in equilibrium all 
the conditions for the equilibrium of rigid bodies ; since, if the 
system of cords became rigid, it would still, with greater rea- 
son, be in equilibrium. 

Prin. III. — The conditions of equilibrium cannot be applied, 
of course, unless the system can be considered a free body, 
which is allowable only when we conceive to be put in, at the 
points of support or fastening, the reactions (upon the cord) 
oi those points and the supports removed. These reactions 
having been put in, then consider the case in Fig. 45 in one 
plane. If we take any point, p, on the cord as a centre 
of moments, knowing that the resultant P, of the forces P ir 
jP a , and P % , situated on one side oip, must act along the cord 

through p (by Prin. 1), therefore 
we have P x a y — P 7 a a — P^ 
v^ = P x zero = 0, and (equally 
P * well) P.a, - P 5 a B - P 4 a 4 = 0. 
That is, in a system of cords in 
FlG> 45> equilibrium in a plane, if a centre 

of moments be taken on the cord, the algebraic sum of the mo- 




STATICS OF FLEXIBLE CORDS. 



43 



merits of those forces situated on one side {either) of this point 
will equal zero. 

41. The Pulley. — A cord in equilibrium over a pulley whose 
axle is smooth has the same tension on both sides ; for, Fig. 46, 





Fig. .46. 



Fig. 47. 



considering the pulley and its portion of cord free 2(Pa) = 
about the centre of axle gives P'r = Pr, i.e., P' = P = ten- 
sion in the cord. Hence the pressure P at the axle bisects 
the angle a, and therefore if a weighted pulley rides upon a 
cord ABC, Fig. 47, its position of equilibrium, B, may be 
found by cutting the vertical through A by an arc of radius 
CD = length of cord, and centre at (7, and drawing a horizon- 
tal through the middle of AD to cut CD in B. A smooth 
ring would serve as well as the pulley ; this would be a slip- 
knot. 



MW////// 
c 



42. If three cords meet at afxed knot, and are in equilib- 
rium, the tension in any one is the equal and 
opposite of the resultant of those in the other 
two. 

43. Tackle. — If a cord is continuous over a 
number of sheaves in blocks forming a tackle, 
neglecting the weight of the cord and blocks and 
friction of any sort, we may easily find the ratio 
between the cord-tension P and the weight to be 
sustained. E.g., Fig. 48, regarding all the straight 
cords as vertical and considering the block B 
free, we have, Fig. 49 (from 2Y= 0), 4P — G 

G 




= 0,.-.P 



The stress on the support C will = 5P. 



44 



MECHANICS OF ENGINEERING. 




44. Weights Suspended by Fixed Knots. — Given all the geo- 
metrical elements in Fig. 50, and 
one weight, £?,; required the re- 
maining weights and the forces 

ny n & & 

H , Vq, H n and V m at the points 
of support, that equilibrium may 
obtain. H and V are the hori- 
zontal and vertical components of 
the tension in the cord at 0\ 
similarly H n and V n those at n. There are n-{-2 unknowns. 
From Prin. II we have 2X = 0, and 2T = 0; i.e., H - H n 
= 0, and [G x + # 3 + ...]-[ Vo + Vn\ = 0. While from 
Prin. III., taking the successive knots, 1, 2. etc., as centres of 
moments, we have 

— Vox, + H,y x = 0, 

— V x, + H y, -f Gfa - x,) = 0, 

— V x 3 + B y z + G t {x a - x,) + G 2 (x 3 — x t ) = 0, 

etc., for n knots. 

Thus we have n -\- 2 independent equations, a sufficient 
number, and they are all of the first degree (with reference to 
the unknowns), and easily solved. As a special solution, we 
may, by § 42, resolve G x in the directions of the first and sec- 
ond cord-segments, and obtain their tensions by a parallelogram 
of forces ; then at the second knot, knowing the tension in the 
second segment, we may find that in the third and G 3 in like 
manner, and so on. Of course H and V are components of 
the tension in the first segment, H n and V n of that in the 
last. 

45. The converse of the problem in § 44, viz., given the 
weights G„ etc., x n and y n , the lengths a, h, c, etc.; required 
H , V , H n , V n , and the co-ordinates x v y v a? 2 , y^ etc., of the 
fixed knots when equilibrium exists, contains 2n -\- 2 un- 
knowns. Statics furnishes n -f- 2 equations (already given in 
§44); while geometry gives the other n equations, one for 
each cord-segment, viz., x* -f- y? = a 2 ; (a? s — x^f -\- (y 2 — y^ 
— ¥ ; etc. 



STATICS OF FLEXIBLE CORDS. 



45 



!Y 



.. x _ m/m 




However, most of these 2n -\- 2 equations are of the second 
degree ; hence in the general case they cannot be solved. 

46. Loaded Cord as Parabola. — If the weights are equal and 
infinitely small, and are intended to be uniformly spaced 
along the horizontal, when equilib- 
rium obtains, the cord having no 
weight, it will form a parabola. Let 
q = weight of loads per horizontal 
linear unit, O be the vertex of the 
curve in which the cord hangs, and 
m any point. We may consider 
the portion Om as a free body, if 
the reactions of the contiguous portions of the cord are put in r 
H Q and T, and these (from Prin. I.) must act along the tangents 
to the curve at O and m, respectively ; i.e., H is horizontal,. 

dy 
and T makes some angle cp (whose tangent = -/-, etc.) with 

the axis X. Applying Prin. II., 

fix 

2X = gives T cos (p-B =0', i.e., Tj 8 = B ; . . (1) 
2 Y= gives T sin cp - qx = ; i.e., T^F — qXm > # ( 2 ) 
Dividing (2) by (1), member by member, we have ^- = %= - 
r . dy — -jj-xdx, the differential equation of the curve; 
v = ~Yr / xdx = 



X 

. =r ; or x 



rr • o » U1 " — Vi * ne equation of a 

; ~" Jj-o £ q 

parabola whose vertex is at and axis vertical. 



Note. 



-The same result, ~ = ~, mav be obtained by considering that 
dx JIq 

we have here (Prin. II.) a free rigid body acted on 

by three forces, T, Hq, and R = qx, acting verti- 

■" cally through the middle of the abscissa x\ the 

resultant of Hq and R must be equal and oppo- 

X site to T, Fig. 52. . \ tan <p = ^-, or % = fL 

Uq ax Hq, 

Evidently also the tangent-line bisects the ab- 
scissa x. 




'1 



-46 MECHANICS OF ENGINEERING. 

47. Problem under § 46. [Case of a suspension-bridge in 
which the suspension-rods are vertical, the weight of roadway 
is uniform per horizontal foot, and large compared with that 
of the cable and rods. Here the roadway is the only load : it 
is generally furnished with a stiffening truss to avoid deforma- 
tion under passing loads.] — Given the span = 2b, Fig. 53, 

Yj v f 71 * ne deflection = a, and the rate of loading 

j^V = q lbs. per horizontal foot ; required the 
tension in the cable at 0, also at m ; and 
"h~~ i i i i ~t I IT '"" ~* *^ e l en &th of cable needed. From the 
Fig. 53. equation of the parabola qx 2 = 2H y, put- 

ting x = b and y = a, we have JI = qb* -i- 2a — the tension 
at 0. From 2Y= we have V x — qb, while 2X = gives 

H, = B ; .'. the tension at m = VHf + V^ ~[qb tftf+Jf]. 

The semi-length, Om , of cable (from p. 88, Todhunter's In- 
tegral Calculus) is (letting n denote U -i- 2q) 

Om = Vna + a? -f- n • 1°©* [( ^ a + ^ n + a ) + ^ n \ 

48. The Catenary. — A flexible, inextensible cord or chain, of 
uniform weight per unit of length, hung at two points, and 
supporting its own weight alone, forms a curve called the 
catenary. Let the tension H at the lowest point or vertex be 
represented (for algebraic convenience) by the weight of an 
imaginary length, c, of similar cord weighing q lbs. per unit 
of length, i.e., H = qc\ an actual portion of the cord, of 
length 8, weighs qs lbs. Fig. 54 shows as free and in equilib- 

j rium a portion of the curve of any 
length s, reckoning from the 
vertex. Required the equation of 
the curve. The load is uniformly 
spaced along the curve, and not 
horizontally, as in §§ 46 and 47. 

fig. 54. 2 Y = gives 2-~ = qs ; while 

(i i* 
2X = gives Tj- = qc. Hence, by division, cdy = sdx, and 

squaring, cfdy* = s'dx* (1) 




dx 



STATICS OF FLEXIBLE CORDS. 47 

Put dy* = ds 2 — dx*, and we have, after solving for dx 
cds r* s ds 



- -''*=«/wh;=€ lo ^ s + Vs2+ ^ 



Vs* + c*' t/o Vs* + c 



and x = c . log e [(5 + iV + c 2 ) -f- <?], . . . (2) 

a relation between the horizontal abscissa and length of curve. 
Again, in eq. (1) put dx* = ds* — dy 2 , and solve for dy. 

This gives dy = f== = \ ■ %%'if Therefore 

V = i£(o' + sTW + «") - i\j(<? + sj, and finally 

y = VT+? -C (3) 

Clearing of radicals and solving for o, we have 

c = (s*-y*) + 2y. ..... (4) 

Example. — A 40-foot chain weighs 240 lbs., and is so hung 
from two points at the same level that the deflection is 10 
feet. Here, for s = 20 ft., y= 10; hence eq. (4) gives the 
parameter, e = (400 — 100) -f- 20 = 15 feet, q = 240 -=- 40 
= 6 lbs. per foot. .*. the tension at the middle is S = qc 
= 6 x 15 = 90 lbs.; while the greatest tension is at either 
support and = VW -f 120 2 = 150 lbs. 

Knowing c = 15 feet, and putting s = 20 feet = half 
length of chain, we may compute the corresponding value of 
x from eq. (2) ; this will be the half-span [log e m = 2.30258 
X (common log m)]. To derive s in terms of x, transform 
eq. (2) in the sense in which n = log e m may be transformed 
into £ n = m, clear of radicals, and solve for s, which gives 

S =ic^y - f -3 ] (4) 

Again, eliminate s from (2) by substitution from (3), trans- 
form as above, clear of radicals, and solve for y -f- <?, whence 

r + a = *,[/+.-=], (5) 



48 MECHANICS OF ENGINEERING. 

which is the equation of a catenary with axes as in Fig. 54. 
If the horizontal axis be taken a distance == c below the ver- 
tex, the new ordinate y' = y -f- c, while x remains the same; 
the last equation is simplified. 

If the span and length of chain are given, or if the span 
and deflection are given, c can be determined from (4) or (5) 
only by successive assumptions and approximations. 



PART II -DYNAMICS. 



CHAPTER I. 

RECTILINEAR MOTION OF A MATERIAL POINT. 

49. Uniform Motion implies that the moving point passes 
over equal distances in equal times ; variable motion, that un- 
equal distances are" passed over in equal times. In uniform 
motion the distance passed over in a unit of time, as one sec- 
ond, is called the velocity (= v), which may also be obtained 
by dividing the length of any portion (= s) of the path by 
the time (= t) taken to describe that portion, however small or 
great ; in variable motion, however, the velocity varies from 
point to point, its value at any point being expressed as the 
quotient of ds (an infinitely small distance containing the 
given point) by dt (the infinitely small portion of time in 
which ds is described). 

49#. By acceleration is meant the rate at which the velocity 
of a variable motion is changing at any point, and may be a 
uniform acceleration, in which case it equals the total change 
of velocity between any two points, however far apart, divided 
by the time of passage ; or a variable acceleration, having a 
different value at every point, this value then being obtained 
by dividing the velocity-increment, dv, or gain of velocity 
in passing from the given point to one infinitely near to it, by 
dt, the time occupied in acquiring the gain. (Acceleration 
must be understood in an algebraic sense, a negative accelera- 
tion implying a decreasing velocity, or else that the velocity in 
a negative direction is increasing.) The foregoing applies to 
motion in a path or line of any form whatever, the distances 
mentioned being portions of the path, and therefore measured 
along the path. 
4 



50 MECHANICS OF ENGINEERING. 

50. Rectilinear Motion, or motion in a straight line. — The 
general relations of the quantities involved may be thus stated 
(see Fig. 55) : Let v = velocity of the body at any instant ; 

-S 0< s —»dsds_ +S * hen dv = § ain of velocity 

1 \ T" -1 ? in an instant of time dt. Let 



Fig. 55. 



dndt> '» t = time elapsed since the 
body left a given fixed point, 
which will be taken as an origin, O. Let s = distance (-(- or 
— ) of the body, at any instant, from the origin O ; then ds = 
distance traversed in a time dt. Let^> = acceleration = rate 
at which v is increasing at any instant. All these may be 
variable ; and t is taken as the independent variable, i.e., time 
is conceived to elapse by equal small increments, each == dt ; 
hence two consecutive ^s will not in general be equal, their 
difference being called d?s. Evidently d 2 t is = zero, i.e., dt is 
constant. 

Since -j- = number of instants in one second, the velocity at 

any instant (i.e., the distance which would be described at that 

rate in one second) is v = ds . -j- ; .*. v — -j-. . . . . (I.) 

! / /ds\ a Ti s\ 

Similarly, p = dv . ---, and ^since dv = dy dt J = -^y 

dv d*s 

■'•&~-4 = de (IL > 

Eliminating dt, we have also vdv =pds (HI.) 

These are the fundamental differential formulae of rectilinear 
motion (for curvilinear motion we have these and some in ad- 
dition) as far as kinematics, i.e., as far as space and time, is 
concerned. The consideration of the mass of the material 
point and the forces acting upon it will give still another rela- 
tion (see § 55). 

51. Rectilinear Motion due to Gravity. — If a material point 
fall freely in vacuo, no initial direction other than vertical 
having been given to its motion, many experiments have 






EECTILINEAR MOTION OF A MATERIAL POINT. 51 

shown that this is a uniformly accelerated rectilinear motion 
in a vertical line having an acceleration (called the accelera- 
tion of gravity) equal to 32.2 feet per square second, or 9.81 
metres per square second; i.e., the velocity increases at this 
constant rate in a downward direction, or decreases in an up- 
ward direction. 

[Note. — By " square second " it is meant to lay stress on the fact that an 
acceleration (being = cPs -*• dP) is in quality equal to one dimension of 
length divided by two dimensions of time. E.g., if instead of using the 
foot and second as units of space and time we use the foot and the minute, 
g will = 32.2 X 3600; whereas a velocity of say six feet per second would 
= 6 X 60 feet per minute. The value of g = 32.2 implies the units foot 
and second, and is sufficiently exact for practical purposes.] 

52. Free Fall in Vacuo.— Fig. 56. Let the body start at 
with an initial downward velocity = c. The accelera- _s 
tion is constant and = -f- g. Reckoning both time and 
distance (+ downwards) from 0, required the values of *° 
the variables s and v after any time t. From eq. (II.), "\ o 
§ 50, we have -f- g = dv -f- dt ; .;. dv = gdt, in which the * s 
two variables are separated. 

o dv = 9Jo dt > i - e '? \_ C V = 9\J\ OV V^—C = V 

gt—0; and finally, v = c + gt (1) fig. 56. 

(Notice the correspondence of the limits in the foregoing 
operation ; when t = 0, v = + c.) 

From eq. (I.), § 50, v = ds -^- dt ; /. substituting from (1), 
ds = (c -f- gt)dt, in which the two variables s and t are sepa- 
rated. 

/s pt pt rs p r-t £2 

ds = cj dt + gJ Q tdt;i.e.,[_ Q S=c) L t+g\_ o -, 

or s = ct + igf (2) 

Again, eq. (III.), § 50, vdv = gds, in which the variables v 
and s are already separated. 

•'• S vdv = g J>* ds * or [V = g\j ; i-e., K*> a - O = 9*, 



am 



52 MECHANICS OF ENGINEERING. 

If the initial velocity = zero, i.e., if the body falls from rest, 



& 

eq. (3) gives s = <^and<u = Vtyh. [From the frequent re- 



currence of these forms, especially in hydraulics, ^-is called the 

"height due to the velocity v," i.e., the vertical height through 
which the body must fall from rest to acquire the velocity v ; 
while, conversely, ^2gh is called the velocity due to the height 
or head A.] 

By eliminating g between (1) and (3), we may derive another 
formula between three variables, s, v, and t, viz., 

s = i(c + v)t (4) 

53. Upward Throw. — If the initial velocity were in an up- 
ward direction in Fig. 56 we might call it — c, and introduce it 
with a negative sign in equations (1) to (4), just derived; but 
for variety let us call the upward direction -(-, in which case 
an upward initial velocity would = -f- c, while the acceleration 
= — g, constant, as before. (The motion is supposed confined 
within such a small range that g does not sensibly vary.) Fig. 

I 57. From^> = dv -~ dt we have dv = — gdt and 

J dv = — gj o dt ; .\ v — c = — gt ; or v = c — gt. (l)a 



ill 



-Q 

From v = ds -f- dt, ds == cdt — gtdt, 



s*s r*t nt 

i.e., J Q ds = cJ o dt - gj tdt ; or s=ct- \gt\ (2)a 

i T i nv r*s 

— S vdv = pds gives / vdv = — gJ Q ds, whence 



Fig. 57 



%(v 2 - c') = - gs, or finally, * = . . (3)cs 

And by eliminating g from (l)a and (3)#, 

« = *(* + «)* (*)* 

The following is now easily verified from these equations : 
the body passes the origin again (s = 0) with a velocity = — c, 
after a lapse of time = 2c -5- <?. The body comes to rest (for 






RECTILINEAR MOTION OF A MATERIAL POINT. 53 

an instant) (put v = 0) after a time = o ~ g, and at a distance 
s = & -T- 2g (" height due to velocity c"') from O. For tf > 
■c -7- g, v is negative, showing a downward motion ; for t > 
2c -7- g, s is negative, i.e., the body is below the starting-point 
while the rate of change of v is constant and = — g at all 
points. 

54. Newton's Laws. — As showing the relations existing in 
general between the motion of a material point and the actions 
(forces) of other bodies upon it, experience furnishes the fol- 
lowing three laws or statements as a basis for dynamics : 

(1) A material point under no forces, or under balanced 
forces, remains in a state of rest or of uniform motion in a 
right line. (This property is often called Inertia.) 

(2) If the forces acting on a material point are unbalanced, 
an acceleration of motion is produced, proportional to the re- 
sultant force and in its direction. 

(3) Every action (force) of one body on another is always 
accompanied by an equal, opposite, and simultaneous reaction. 
(This was interpreted in § 3.) 

As all bodies are made up of material points, the results ob- 
tained in Dynamics of a Material Point serve as a basis for the 
Dynamics of a Rigid Body, of Liquids, and of Gases. 

55. Mass. — If a body is to continue moving in a right line, 
the resultant force P at all instants must be directed along that 
line (otherwise it would have a component deflecting the body 
from its straight course). 

In accordance with Newton's second law, denoting by^> the 
acceleration produced by the resultant force (G being the 
body's weight), we must have the proportion P : G : : jp : g ; 
i.e., 

P = -.p , or P = Mp. . . (IY.) 

Eq. IY. and (I.), (II.), (DDL) of § 50 are the fundamental 
equations of Dynamics. Since the quotient G ■— g is invaria- 



54 MECHANICS OF ENGINEERING. 

ble, wherever the body be moved on the earth's surface (G and 
g changing in the same ratio), it will be used as the measure 
of the mass M or quantity of matter in the body. In this way 
it will frequently happen that the quantities G and g will ap- 
pear in problems where the weight of the body, i.e., the force 
of the earth's attraction upon it, and the acceleration of gravity 
have no direct connection with the circumstances. No name 
will be given to the unit of mass, it being always understood 
that the fraction G -f- g will be put for M before any numeri- 
cal substitution is made. From (IY.) we have, in words, 

j accelerating force = mass X acceleration; 

\ also, acceleration — accelerating force -r- mass. 

56. Uniformly Accelerated Motion. — If the resultant force is 
constant as time elapses, the acceleration must be constant (from 
eq. (IV.), since of course M is constant) and = P -f- M. The 
motion therefore will be uniformly accelerated, and we have 
only to substitute +J9 (constant) for g in eqs. (1) to (4) of 
§ 52 for the equations of this motion, the initial velocity being 
= c (in the line of the force). 

v = o+pt ... (1); s = ct + ipf; ... (2) 

S = ^T>' ' " (3) ' ™ ds = H G + v)t ... (4) 

If the force is in a negative direction, the acceleration will 
be negative, and may be called a retardation/ the initial veloc- 
ity should be made negative if its direction requires it. 

57. Examples of Unif. Ace. Motion. — Example 1. Fig. 58. 
A small block whose weight is \ lb. has already described a 
-S c P M v B distance Ao == 48 inches over a 

A smooth -^7» ^r ZJEfa^-. j> sm ooth portion of a horizontal 
FlG . 58. table in two seconds ; at it en- 

counters a rough portion, and a consequent constant friction of 
2 oz. Kequired the distance described beyond 0, and the time 
occupied in coming to rest. Since we shall use 32.2 for g r 
times must be in seconds, and distances in feet ; as to the unit 



RECTILINEAR MOTION OF A MATERIAL POINT. 55 

of force, as that is still arbitrary, say ounces. Since A was 
smooth, it must have been described with a uniform motion 
(the resistance of the air being neglected); hence with a veloc- 
ity = 4 ft. -7- 2 sec. = 2 ft. per sec. The initial velocity for 
the retarded motion, then, is c = -f- 2 at 0. At any point be- 
yond the acceleration = force -J- mass = (— 2 oz.) -r- (8 oz. 
-r- 32.2) = — 8.05 ft. per square second, i.e., p = — 8.05 = 
constant ; hence the motion is uniformly accelerated (retarded 
here), and we may use the formulae of § 56 with c = -f- 2, p = 
— 8.05. At the end of the motion v must be zero, and the 
corresponding values of s and t may be found by putting v = 
in equations (3) and (1), and solving for s and t respectively : 
thus from (3), * = \ (- 4) -f- (— 8.05), i.e., s = 0.248 +, which 
must be feet ; while from (1), t = (— 2) -i- (— 8.05) = 0.248 +, 
which must be seconds. 

Example 2. (Algebraic.) — Fig. 59. The two masses M x == 
G 1 -T- g and M= G -f- g are connected by a flexible, inexten- 
sible cord. Table smooth. Required the acceleration common 
to the two rectilinear motions, and the tension in the string S y 



i^s 




ispf 

Fig. 59. Fig. 60. 

there being no friction under G l9 none at the pulley, and no 
mass in the latter or in the cord. At any instant of the mo- 
tion consider G x free (Fig. 60), N being the pressure of the 
table against G x , Since the motion is in a horizontal right line 
J£(vert. compons.)= 0, i.e., N — G x = 0, which determines J¥~. 
S, the only horizontal force (and resultant of all the forces) = 
Mj>, i.e., 

8=G lP ~g (1) 

At the same instant of the motion consider G free (Fig. 61); 
the tension in the cord is the same value as above = S. The 
accelerating force is G — 8, and 

.". = mass X ace, or G — S = (G -f- g)p. . (2) 



56 MECHANICS OF ENGINEERING. 



* 
V 



| s From equations (1) and (2) we obtain p = (Gg) -f- 

(G -f- 6^) = a constant ; hence each motion is uniformly 

r 1 S accelerated, and we may employ equations (1) to (4) of 

§ 56 to find the velocity and distance from the starting- 



H s 






points, at the end of any assigned time t, or vice versa. 
' The initial velocity must be known, and may be zero. 
Also, from (1) and (2) of this article, 

S = (GG,) + (G+ G,) = constant. 

Example 3. — A body of 2f (short) tons weight is acted on 
during -J minute by a constant force P. It had previously de- 
scribed 316f yards in 180 seconds under no force ; and subse- 
quently, under no force, describes 9900 inches in ^ of an hour. 
Eequired the value of P. Ans. P = 22.1 lbs. 

Example 4. — A mass of 1 ton having an initial velocity of 48 
inches per second, is acted on for J minute by a force of 400 
avoirdupois ounces. Required the final velocity. 

Ans. 10.037 ft. per sec. 

Example 5. — Initial velocity, 60 feet per second ; mass weighs 
0.30 of a ton. A resistance of 112^ lbs. retards it for -^ of 
a minute. Required the distance passed over during this time. 

Ans. 286.8 feet. 

Example 6. — Required the time in which a force of 600 avoir- 
dupois ounces will increase the velocity of a mass weighing 1J 
tons from 480 feet per minute to 240 inches per second. 

Ans. 30 seconds. 

Example 7. — What distance is passed over by a mass of (0.6) 
tons weight during the overcoming of a constant resistance 
(friction), if its velocity, initially 144 inches per sec, is reduced 
to zero in 8 seconds. Required, also, the friction. 

Ans. 48 ft. and 55 lbs. 

Example 8. — Before the action of a force (value required) a 
body of 11 tons had described uniformly 950 ft. in 12 minutes. 
Afterwards it describes 1650 feet uniformly in 180 seconds. 
The force acts 30 seconds. P = % Ans. P = 178 lbs. 



RECTILINEAR MOTION OF A MATERIAL POINT. 



57 



58. Graphic Representations. Unif. Ace. Motion. — With the 
initial velocity = 0, the equations of § 56 become 

v=pt, (1) s = ipt% (2) 

s = v* -^ 2p, . . . (3) and s — ivt (4) 

Eqs. (1), (2), and (3) contain each two variables, which may 
graphically be laid oft" to scale as co-ordinates and thus give a 
curve corresponding to the equation. Thus, Fig. 62, in (I.), we 

MQ- 
h /? 

«i_^^_Jj - h^r_ l J7 

(in.) 

have a right line representing eq. (I.) ; in (II.), a parabola with 
axis parallel to s, and vertex at the origin for eq. (2) ; also a 
parabola similarly situated for eq. (3). Eq. (4) contains three 
variables, s, v, and t. This relation can be shown in (I.), s be- 
ing represented by the area of the shaded triangle = %vt. 
(II.) and (III.) have this advantage, that the axis OS may be 
made the actual path of the body. [Let the student determine 
how the origin shall be moved in each case to meet the supposi- 
tion of an initial velocity = -|- c or — <?.] 





59. Variably Accelerated Motions. — We here restate the equa- 
tions 

Vz= dt * " ^ ; p = ~di = df ' ' ( IL ) ; vdv =1>ds ' ' ( IIL ) ; 
and resultant force 

= P = M P , (IV.); 

which are the only ones for general use in rectilinear motion. 

Problem 1. — In pulling a mass M along a smooth, horizon- 
tal table, by a horizontal cord, the tension is so varied that 
s = 4cf (not a homogeneous equation ; the units are, say, the 
foot and second). Required by what law the tension varies. 



58 MECHANICS OF ENGINEERING. 

ds d(±f) _ 2 , _ x <J(12*») 

From (I.) . = dt = -y = Itf ; from (II), j, = ^ = 

24£; and (IV.) the tension = P = Mp = 24:2ft, i.e., varies 
directly as the time. 

Problem 2. " Harmonic Motion," Fig. 63. — A small block 






-s 9 _r— *-».« 

J t^^ 1 , 

Fig. 63 

on a smooth horizontal table is attached to two horizontal 
elastic cords (and they to pegs) in such a way that when the 
block is at 0, each cord is straight but not tense ; in any other 
position, as m, one cord is tense, the other slack. The cords 
are alike in every respect, and, as with springs, the tension 
varies directly with the elongation (= s in figure). If for an 
elongation s x the tension is T l9 then for any elongation s it is 
T = T x s -r- s x . The acceleration at any point m, then, is 
p = — (T -r- M) — — (T x s -v- J/sJ, which for brevity put 
p ±= — as, a being a constant. Required the equations of 
motion, the initial velocity being = -f- c, at O. From eq. (III.) 

vdv = — asds ; .*. / vdv = — a J sds, 

i.e., i(v 2 — c 2 ) = — J-as 2 ; or, v* = c* — as 2 . , (1) 
From (I.), dt = ds ~ v; hence from (1), 



or 



t = ~j=- / ■ r ±—= =d= = —= sin- i( _L? I 



1 . , /* Va\ 

—-^sin- 1 ). », 

Va v ^ / 



= — ^sin 



RECTILINEAR MOTION OF A MATERIAL POINT. 59 



Inverting (2), we have s = (c -r- Va) sin (t Va), . . . (3) 
Again, by differentiating (3), see (I.), v = c cos (t Va) (4) 
Differentiating (4), see (IL)>i> = — c Va sin (t Va). . . (5) 

These are the relations required, but the peculiar property 
of the motion is made apparent by inquiring the time of pass- 
ing from to a state of rest ; i.e., put v = in equation (4),. 
we obtain t = \n -f- Va, or J-7T — Va, or f -it -f- Va, and so on, 
while the corresponding values of s (from equation (3)), are 
-f- (c -r- Va), — (c ~ Va), -f- (o -r- Va), and so on. This shows 
that the body vibrates equally on both sides of in a cycle or 
period whose duration = 2^r -^- Va, and is independent of the 
initial velocity given it at 0. Each time it passes the 
velocity is either -f- c, or — c, the acceleration = 0, and the 
time since the start is = nn -f- Va, in which ^ is any whole 
number. At the extreme point p = =F c Va, from eq. (5). 
If then a different amplitude be given to the oscillation by 
changing c, the duration of the period is still the same, i.e.,. 
the vibration is isochronal. The motion of an ordinary pen- 
dulum is nearly, that of a cycloidal pendulum exactly, harmonic. 

If the crank-pin of a reciprocating engine moved uniformly 

in its circular path, the piston would have a harmonic motion 

if the connecting-rod were infinitely long, or if the design in. 

-< — -2-r -— — > 

JILJH?., 



~WTM 



Fig. 64. 




Fig. 64 were used. (Let the student prove this from eq. (3).) 
Let 2r = length of stroke, and c = the uniform velocity of the 
crank-pin, and M = mass of the piston and rod AB. Then 
the velocity of M at mid-stroke must = c, at the dead-points, 
zero; its acceleration at mid-stroke zero; at the dead-points 
the ace. = c Va, and s = r — c -r- Va (from eq. (3)) ; .*. Va 
= c ~ r, and the ace. at a dead point (the maximum ace.) 



60 



MECHANICS OF ENGINEERING 






Fig. 65. 



= c a -T- r. Hence on account of the acceleration (or retarda- 
tion) of M in the neighborhood of a dead-point a pressure will 
be exerted on the crank-pin, equal' to mass X ace. = Mc~ -=- r 
at those points, independently of the force transmitted due to 
steam-pressure on the piston-head, and makes the resultant 
pressure on the pin at C smaller, and at D larger than it would 
be if the "inertia" of the piston and rod were not thus taken 
into account. "We may prove this also by the free-body method, 
considering AB free immediately after passing the dead-point 

C, neglecting all friction. See Fig. 
65. The forces acting are : G, the 
weight ; j¥, the pressures of the 
guides ; P, the known effective steam- 
pressure on piston-head ; and P\ the unknown pressure of 
crank-pin on side of slot. There is no change of motion ver- 
tically ; .-. JUT + JSf— G = 0, and the resultant force is P — P' 
= mass X accel. = Mc* ~- r, hence P' = P — M& — r. 
Similarly at the other dead-point we would obtain P' = P -f- 
M& -r- r. In high-speed engines with heavy pistons, etc., 
M& -=- r is no small item. [The upper half-re vol., alone, is 
here considered.] 

Problem 3. — Supposing the earth at rest and the resistance 
of the air to he null, a body is given an initial upward vertical 
velocity = c. Required the velocity at any distance s from 
the centre of the earth, whose attraction varies in- 
versely as the square of the distance s. 

See Fig. 66. — The attraction on the body at the 
surface of the earth where s = r, the radius, is its 
weight G ; at any point m it will be P = G(r* -f- s 2 ), 
while its mass = G -f- g. 

Hence the acceleration at m = j? = ( — P) ~- M 
— — 9{ r<2 ~+~ <* 2 )- Take equation III., vdv = pds, 
and we have 

vdv = — c/r's ~ 2 ds: .*. 







ds 



I vdv = — gr 2 I 

i.e., i(v> - c 2 ) = 



or, 



[> ! =-^D4 



y 



a) 



RECTILINEAR MOTION OF A MATERIAL POINT. 



61 



Evidently v decreases, as it should. Now inquire how small 
a value c may have that the body shall never return; i.e., 
that v shall not = until s = oo. Put v = and s = oo in 
(1) and solve for c ; and we have 



c = 4/2^r = V2 X 32.2 X 21000000, 

= about 36800 ft. per sec. or nearly 7 miles per sec. Con- 
versely, if a body be allowed to fall, from rest, toward the 
earth, the velocity with which it would strike the surface 
would be less than seven miles per second through whatever 
distance it may have. fallen. 

If a body were allowed to fall through a straight opening in 
the earth passing through the centre, the motion would be har- 
monic, since the attraction and consequent acceleration now 
vary directly with the distance from the centre. See Prob. 2. 
This supposes the earth homogeneous. 

Problem 4. — Steam working expansively and raising a weight. 
Fig. 67. — A piston works without 
friction in a vertical cylinder. Let 
S = total steam-pressure on the 
under side of the piston ; the weight 
G, of the mass G -— g (which in- 
cludes the piston itself) and an 
atmospheric pressure = A, con- 
stitute a constant back-pressure. 
Through the portion OB = s t , of Fig. 67. 

the stroke, Sis constant = S„ while beyond B, boiler com- 
munication being " cut off," S diminishes with Boyle's law, i.e., 
in this case, for any point above B, we have, neglecting the 
" clearance", F being the cross-section of the cylinder, 



N/" 

/ 

/ 




g* r 

1A 

mm 






V7///>, 




\ 

M 




* y t > <^— - 




: 


3 A 
IP 




O . 


' 




SiS.r.Fs,: Fs; or S = S, 



s. 



Pull length of stroke = ON = s n . Given, then, the forces 
Si and A y the distances s t and s n , and the velocities at O and 
at iVboth = (i.e., the mass M = G -f- g is to start from rest at 
O, and to come to rest at iV), required the proper weight G to 



€2 MECHANICS OF ENGINEERING. 

fulfil these conditions, S varying as already stated. The accel- 
eration at any point will be 

p=[S-A-GF\ + M. .... (1) 

Hence (eq. III.) Mvdv = [S — A — G]ds, and ,\ for the 

whole stroke 

Mf Q vdv=f Q [S-A- G]ds; i.e., 

z>si p s *ds z*^ /* s, > 

or _ 8 1 s 1 [l + log/f2 = As K +G gn ,. . . (2) 

Since S = S x = constant, from O to B, and variable, = 
S l s 1 -r- *, from i? to JW, we have had to write the summation 



jf 



N 

/Sds in two parts. 



From (2), G becomes known, and .*. Jf also (= G -~ g). 
Required, further, the time occupied in this upward stroke. 
From to B (the point of cut-off) the motion is uniformly 
accelerated, since p is constant (S being = S x in eq. (1) ), 
with the initial velocity zero; hence, from eq. (3), § 56, 
the velocity at B = v 1 = V2 [S x — A — G]s 1 -f- M is known ; 
.-. the time t x = 2s 1 —■ v x becomes known (eq. (4), § 56) of de- 
scribing OB. At any point beyond B the velocity v may be ob- 
tained thus : From (III.) vdv = pds, and eq. (1) we have, 
summing between B and any point above, 

Jf£«k> = SaJT d i-(A + G)£ds ; i.e., 

This gives the relation between the two variables v and s 
anywhere between B and iT; if we solve for -y and insert its 
value in dt =■ ds -±- v, we shall have dt = a function of 5 and 
■dsj which is not integrable. Hence we may resort to approxi- 



RECTILINEAR MOTION OF A MATERIAL POINT. 63 

mate methods for the time from B to JV. Divide the space 
BN into an uneven number of equal parts, say five ; the dis- 
tances of the points of division from will be s t , s„ s s1 s t , s„ 
and s n . For these values of s compute (from above equation) 
v x (already known), v» v„ i> 4 , v„ and v n (known to be zero). To 
the first four spaces apply Simpson's Rule, and we have the 
time from B to the end of s„ 

r 6 r* <k « 6 -«.ri , 4 , 2 , 4 n 

while regarding the motion from 5 to iV as uniformly retarded 
(approximately) with initial velocity = v 6 and the final = zero, 
we have (eq. (4), § 56), 



r-N 

t — 2(s n — * 5 ) -7- v 6 . 

I— 6 



By adding the three times now found we have the whole time 
of ascent. In Fig. 67 the dotted curve on the left shows by 
horizontal ordinates the variation in the velocity as the distance 
s increases ; similarly on the right are ordinates showing the 
variation of /S. The point E, where the velocity is a maximum 
= v m , may be found by putting p = 0, i.e., f or S = A -{- G, 
the acelerating force being = 0, see eq. (1). Below E the ac- 
celerating force, and consequently the acceleration, is positive ; 
above, negative (i.e., the back-pressure exceeds the steam- 
pressure). The horizontal ordinates between the line HE'KL 
and the right line BT&re proportional to the accelerating force. 
If by condensation of the steam a vacuum is produced be- 
low the piston on its arrival at JV, the accelerating force is 
downward and = A -f- G. [Let the student determine how 
the detail of this problem would be changed, if the cylinder 
were horizontal instead of vertical.] 

60. Direct Central Impact. — Suppose two masses M x and M % 
to be moving in the same right line so that their distance apart 
continually diminishes, and that when the collision or impact 
takes place the line of action of the mutual pressure coincides 
with the line joining their centres of gravity, or centres of 



64 MECHANICS OF ENGINEERING. 

mass, as they may be called in this connection. This is called 
a direct central impact, and the motion of each mass is varia- 
bly accelerated and rectilinear during their contact, the only 
force being the pressure of the other body. The whole mass 
of each body will be considered concentrated in the centre of 
mass, on the supposition that all its particles undergo simul- 
taneously the same change of motion in parallel directions. 
(This is not strictly true ; the effect of the pressure being 
gradually felt, and transmitted in vibrations. These vibrations 
endure to some extent after the impact.) When the centres 
of mass cease to approach each other the pressure between the 
bodies is a maximum and the bodies have a common velocity ; 
after this, if any capacity for restitution of form (elasticity) 
exists in either body, the' pressure still continues, but dimin- 
ishes in value gradually to zero, when contact ceases and the 
bodies separate with different velocities. Reckoning the time 
from the first instant of contact, let t' = duration of the first 
period, just mentioned ; t" that of the first -|- the second (resti- 
tution). Fig. 68. Let 3f 1 and M 2 be the masses, and at any 
instant during the contact let v 1 and v, 
be simultaneous values of the velocities 
of the mass-centres respectively (reckon- 
fig. 68. j n g velocities positive toward the right), 

and P the pressure (variable). At any instant the acceleration 
of M 1 is jp x = — (P -T- J/,), while at the same instant that of 
Jf a is jp 2 = -f- (P -f- JQ ; M l being retarded, Jf 2 accelerated, 
in velocity. Hence (eq. II., p = dv -=- dt) we have 

l£ x dv x = — Pdt\ and M t dv, = + Pdt. . . (1) 

Summing all similar terms for the first period of the impact, 
we have (calling the velocities before impact c x and <?„ and the 
common velocity at instant of maximum pressure C) 

mJ c %= - S! pdt ' Le -> m >(° -°.) = - S! pdt i ( 2 > 

Mjjv, = +J^Pdt, i.e., M,(G - c,) = +/W. . (3) 



c^i 




vH 


Ji* 




1 c 


1)1 


1 c 




V? 




p 

M, 






M 2 





RECTILINEAR MOTION OF A MATERIAL POINT. 65 

The two integrals are identical, numerically, term by term, 
since the pressure which at any instant accelerates J/, is nu- 
merically equal to that which retards M x ; hence, though we do 
not know how P varies with the time, we can eliminate the 
definite integral between (2) and (3) and solve for C. If 
the impact is inelastic (i.e., no power of restitution in either 
body, either on account of their total inelasticity or damaging 
effect of the pressure at the surfaces of contact), they continue 
to move with this common velocity, which is therefore their 
final velocity. Solving, we have 

. r _ M' 1 c l + MA 

°- M.+M, W 

Next, supposing that the impact is partially elastic, that the 
bodies are of the same material, and that the summation 

nt" 

1 1 Pdt for the second period of the impact bears a ratio, e, 

to that / Pdt, already used, a ratio peculiar to the material, 

if the impact is not too severe, we have, summing equations 
(1) for the second period (letting V x and V 2 = the velocities 
after impact), 

M > fc V ' dv > = - fJ Pdt > Le -> -*>( V ~ °) - - e S!' Pdt > ( 5 ) 
M, fjdv, = + f'pdt, i.e., Ml V t -0) = +e f Pdt. (6) 

e is called the coefficient of restitution. 

Having determined the value of / Pdt from (2) and (3) in 

terms of the masses and initial velocities, substitute it and that 
of C, from (4), in (5), and we have (for the final velocities) 

V x = [M x c x + lf Ar eM^c-c,)] ~ [M t + MJ; . (7) 

and similarly 

r^lIf^ + MA+eMlc-oM-lM. + M,]. . (8) 
For e=0, i.e., for inelastic impact, V l =- V^= C in eq. (4) ; for 



66 MECHANICS OF ENGINEERING. 

*=1, or elastic impact, (7) and (8) become somewhat simpli- 
fied. 

To determine e experimentally, let a ball (M t ) of the sub- 
stance fall upon a very large slab (Jf 2 ) of the same substance, 
noting both the height of fall K x , and the height of rebound H x . 
Considering M 2 as = oc, with 



c 1 = V 2gh x , V x = — V 2gIT x , and c, = o, 
eq. (7) gives 



- V2gB~ x ,=- eV2gh x \.\e- V H x -~ h x . 

Let the student prove the following from equations (2), (3), 
(5), and (6) : 

(a) For any direct central impact whatever, 

M x c x + M 2 c, = M 1 V 1 + M % V t . 

[The product of a mass by its velocity being sometimes 
called its momentum, this result may be stated thus : 

In any direct central impact the sum of the momenta* before 
impact is equal to that after impact (or at any instant during 
impact). This principle is called the Conservation of Momen- 
tum. The present is only a particular case of a more general 
proposition. 

It may be proved G, eq. (4), is the velocity of the centre of 
gravity of the two masses before impact ; the conservation of 
momentum, then, asserts that this velocity is unchanged by the 
impact, i.e., by the mutual actions of the two bodies.] 

(b) The loss of velocity of M x , and the gain of velocity of 
M„ are twice as great when the impact is elastic as when in- 
elastic. 

(c) If e = 1, and M x = M 2 , then V x = + c 7 , and F 2 = c x . 

Example. — Let Mi and M 2 be perfectly elastic, having weights = 4 and 
5 lbs. respectively, and let d = 10 ft. per sec. and c 2 = — 6 ft. per sec. 
(i. e., before impact M 2 is moving in a direction contrary to that of M^. 
By substituting in eqs. (7) and (8), with e = 1, M t = 4 -*• g, and JT a = 5 ■*- g t 
we have 

7i = I|"4 x 10 + 5 x (- 6) - 5 (lO - (- 6))]= - W ft. per sec. 

Va = ~[~4 x 10 + 5 x (- 6) + 4 (l0 - (- 6))]= + 8.2 ft. per sec. 

as the velocities after impact. Notice their directions, as indicated by their 
signs 



67 



CHAPTEE II. 

"VIRTUAL VELOCITIES." 

61. Definitions. — If a material point is moving in a direction 
not coincident with that of the resultant force acting (as in 
curvilinear motion in the next chapter), and any element of its 
path, ds, projected upon this force ; the length of this projec- 
tion, du, Fig. 69, is called the "Virtual Velocity" of the 
force, since du -4- dt may be considered the veloc- -7 

ity of the force at this instant, just as ds -f- dt is dUyf y^^ 
that of the point. The product of the force by o<0? 
its du will be called its virtual moment, reckoned ds ^4L 
-f- or — according as the direction from to D is 
the same as that of the force or opposite. 



Fig. 69. 



62. Prop. I. — The virtual moment of a force equals the 
algebraic sum of those of its components. Fig. 70. Take the 

p ..-^ direction of ds as an axis X\ let P x and P % 

/ ^^/ be components of P ; a x , a„ and a their 

n/>^^ / angles with X. Then (§16) P cos a = 

C^C ;0\i ? > P 2 -^i cos a A~Pi cos a v Hence P(ds cos a)= 

** ^ ^ jP^s cos <*,) -f- _P 2 (^s cos flr a ). But <£s cos a 

""X = the projection of ds upon P, i.e., = du ; 

Fig. 70. ^ cog ^ _ ^^ etc> . . P^ = P x du x -f- 

P t du r If in Fig. 70 <*, were > 90°, evidently we would 
have Pdu = — P^^j -f- P^du^ i.e., P^i^ would then be 
negative, and (22?, would fall behind 0; hence the definition 
of -J- and — in § 61. For any number of components the 
proof would be similar, and is equally applicable whether they 
are in one plane or not. 

63. Prop. II. — The sum of the virtual moments equals zero, 

for concurrent forces in equilibrium. 



68 MECHANICS OF ENGINEERING. 

(If the forces are balanced, the material point is moving in 
a straight line if moving at all.) The resultant force is zero. 
Hence, from § 62, P x du x -|- P % du^ -f- etc. = 0, having proper 
regard to sign, i.e., 2(Pdu) = 0. 

64. Prop. III. — The sum of the virtual moments equals zero 
for any small displacement or motion of a rigid body in equi- 
librium under non-concurrent forces in a plane; all points of 
the body moving parallel to this plane. (Although the kinds 
of motion of a given rigid body which are consistent with 
balanced non-concurrent forces have not yet been investigated, 
we may imagine any slight motion for the sake of the alge- 
braic relations between the different du's and forces.) 

. First, let the motion be a translation, all 

- — V_ points of the body describing equal parallel 

> \) /:>.- \ lengths = ds. Take Xparallel to ds ; let a iy 

<S&&?* \jr etc., be the angles of the forces with X. 

? V ' p* Then(§35)2(Pcos«) = 0;/.^2(Pcosa) 

\ ; ;v::.^ j s = ; but ds cos a l = du x ; ds cos a 2 = du^ • 
S(fL/ etc. ; .-. 2{Pdu) = 0. Q. E. D. 

4 Secondly, let the motion be a rotation 

Fig. 71. through a small angle dd in the plane of the 

forces about any point O in that plane, Fig. 72. With O as a pole 
let p x be the radius-vector of the point of application of JP„ and 
a 1 its lever-arm from 0; similarly for the 
other forces. In the rotation each point of 
application describes a "small arc, ds x , ds„ 
etc., proportional to p 1? p 2 , etc., since ds l 
= prfd, ds^ = p^dO, etc. From § 36, . 
P x a x -\- etc. == ; but from similar triangles 
ds l : du x :: p x : a x \ .-. a x == p 1 du 1 -^ ds l 
■= du x -r- dd ; similarly a 9 — du a -^ dO, etc. FlG# 72 ' 

Hence we must have \P x du x -j- P 2 du t -\- . . ,~] — dd = 0, i.e.,. 
2(Pdu) = 0. Q. E. D. 

Now since any slight displacement or motion of a body may 
be conceived to be accomplished by a small translation fol- 
lowed by a rotation through a small angle, and since the fore- 




"virtual velocities." 69 

•going deals only with projections of paths, the proposition is 
established and is called the Principle of Virtual Velocities. 
[A similar proof may be used for any slight motion what- 
ever in space when a system of non-concurrent forces is bal- 
anced.] Evidently if the path (ds) of a point of application is 
perpendicular to the force, the virtual velocity (du), and con- 
sequently the virtual moment (Pdu) of the force are zero. 
Hence we may frequently make the displacement of such a 
character in a problem that one or more of the forces may be 
excluded from the summation of virtual moments. 

65. Connecting-Rod by Virtual Velocities. — Let the effective 
steam-pressure P be the means, through the connecting-rod 
and crank (i.e., two links), of raising the weight G very slowly; 
neglect friction and the weight of the links themselves. Con- 
sider AB as free (see (b) in Fig. 73), BG also, at (c); let the 







r^c i_P_ 



dh 

C 



(«•) N. >•"*' (&.) (c) ' 

Fig. 73. 

"small displacements" of both be simtdtaneous small portions 
of their ordinary motion in the apparatus. A has moved to A x 
through dx; B to B„ through ds, a small arc ; G has not 
moved. The forces acting on AB are P (steam-pressure), i\T 
(vertical reaction of guide), and N' and T (the tangential and 
normal components of the crank-pin pressure). Those on BG 
are N' and T (reversed), the weight G, and the oblique pressure 
of bearing P' '. The motion being slow (or rather the accelera- 
tion being small), each of these two systems will be considered as 
balanced. Now put 2(Pdu) = for AB, and we have 

Pdx + #XO + #XO-» = 0. . . (1) 

For the simultaneous and corresponding motion of BG, 

~2£(Pdu) = gives 



p 

A 



70 MECHANICS OF ENGINEERING. 

JST X 0+ Tds- Gdh + P' X = 0, . . (2) 

dh being the vertical projection of G's motion. 

From (1) and (2) we have, easily, Pdx — Gdh = 0, . (3) 

, : /3i which is the same as we might have 

f ^B^\~ h obtained by putting 2(Pdu) = Ofor 

* * * " IGiSrP' the two links together, regarded colr 

* \ lectively as a free body, and describ- 

FlG - 74 - ing a small portion of the motion 

they really have in the mechanism, viz., 

Pdx + lTx 0- Gdh + P' x = 0. . . (4) 

We may therefore announce the — 

66. Generality of the Principle of Virtual Velocities. — If any 

mechanism of flexible inextensible cords, or of rigid bodies 
jointed together, or both, at rest, or in motion with very small 
accelerations, be considered free collectively (or any portion of 
it), and all the external forces put in ; then (disregarding 
mutual frictions) for a small portion of its prescribed motion, 
22(Pdu) must = 0, in which the du, or virtual velocity, of 
each force, P, is the projection of the path of the point of 
application upon the force (the product, Pdu, being -(-or — 
according to § 61). 

67. Example. — In the problem of § 65, having given the 
weight G, required the proper steam-pressure (effective) P to 
hold G in equilibrium, or to raise it uniformly, if already in 
motion, for a given position of the links. That is. Fig. 75 r 
given a, r, c, a, and /?, re- 
quired the ratio dh : dx; for, 
from equation (3), § 65, P 
= G(dh : dx). The projec- A 
tions of dx and ds upon AB dx A, 
will be equal, since AB = FlG - 75 - 

A y B y , and makes an (infinitely) small angle with A t B l9 i.e.,. 
dx cos a—ds cos (/3 — a). Also, dh = (c : r)ds sin /?. 




"virtual velocities." 71 

Eliminating ds, we have, 

dh __ c sin fi cos a # _ c sin yff cos a 

dx ~ r cos (fi — a) ' " " /* cos (/? — «)* 

68. "When the acceleration of the parts of the mechanism is 
not practically zero, 2(Pdu) will not = 0, but a function of 
the masses and velocities to be explained in the chapter on 
Work, Energy, and Power. If friction occurs at moving joints, 
enough " free bodies" should be considered that no free body 
extend beyond such a joint ; it will be found that this friction 
cannot be eliminated in the way in which T and HP were, in 
§65. 

69. Additional Problems ; to be solved by " virtual velocities." 
Problem 1. — Find relations between the forces acting on a 
straight lever in equilibrium ; also, on a bent lever. 

Problem 2. — When an ordinary copying-press is in equilib- 
rium, find the relation between the force applied horizontally 
and tangentially at the circumference of the wheel, and the 
vertical resistance under the screw-shaft. 



72 



MECHANICS OF ENGINEERING. 



CHAPTER III. 



CURVILINEAR MOTION OF A MATERIAL POINT. 




Fig. 76. 



[Motion in a plane, only, will be considered in this chapter.] 

70. Parallelogram of Motions. — It is convenient to regard 
the curvilinear motion of a point in a plane as compounded, or 
made up, of two independent rectilinear motions parallel 
respectively to two co-ordinate axes X and Y, as may be ex- 
plained thus : Fig. 76. Consider the 
drawing-board CD as fixed, and let the 
head of a T-square move from A 
toward B along the edge according to 
any law whatever, while a pencil moves 
from M toward Q along the blade. The 
result is a curved line on the board, whose 
form depends on the character of the 

two X and Y component motions, as they may be called. If 
in a time t x the T-square head has moved an Xdistance = MB, 
and the pencil simultaneously a Y distance = MP, by com- 
pleting the parallelogram on these lines, we obtain i?, the 
position of the point on the board at the end of the time t x * 
Similarly, at the end of the time t x we find the point at It'. 

71. Parallelogram of Velocities. — Let the X and Amotions 
be uniform, required the resulting motion. Fig. 77. Let o„ 
and c y be the constant uniform ^Tand 2^ velocities. Then in 
any time, t, we have x = c x t and y = 
c y t ; whence we have, eliminating t, 
x -T- y = c x -~ c y =- constant, i.e., x is 
proportional to y, i.e., the path is a O^t 
straight line. Laying off OA = c x , U„.£ — % — J 
and AB = c y , B is a point of the path, Fig. 77. 

and OB is the distance described by the point in the first 




CURVILINEAR MOTION OF A MATERIAL POINT. 73 



second. Since by similar triangles OR : x : : OB : c m we 
Lave also OR = OR . t ; hence the resultant motion is uniform, 
and its velocity, OB = c, is the diagonal of the parallelogram 
formed on the two component velocities. 

Corolla?y. — If the resultant motion is curved, the direction 
and velocity of the motion at any point will be given by the 
diagonal formed on the component velocities at that instant. 
The direction of motion is, of course, a tangent to the curve. 

72. Uniformly Accelerated X and Y Motions. — The initial 
velocities of both being zero. Required the resultant motion. 
Fig. 78. From § 56, eq. (2) (both c x andc^ y / ^ 

being = 0), we have x = \p x t and y = ff ^i 

ip y f, whence x -j- y = p x ~ p y = constant, WpCPT / 

and the resultant motion is in a straight jL^ y h>~ L — % 

line. Conceive lines laid off from on X °* x "" 

and Y to represent^ and j? y to scale, and FlG ' 78 ' 

form a parallelogram on them. From similar triangles {OR 
being the distance described in the resultant motion in any 

time t), OR \x\\ ~OB : p x ; .-. OR~= \OBi\ Hence, from the 
form of this last equation, the resultant motion is uniformly 
accelerated, and its acceleration is OB = p, (on the same scale 
as p x and p y ). 

This might be called the parallelogram of accelerations, but 
is really a parallelogram of forces, if we consider that a free 
material point, initially at rest at 0, and acted on simulta- 
neously by constant forces P x and P y (so that p x = P x -f- M 
and^> v = P y ~ M), must begin a uniformly accelerated recti- 
linear motion in the direction of the resultant force, having no 
initial velocity in any direction. 

73. In general, considering the point hitherto spoken of as a 
free material point, under the action of one or more forces, in 
view of the foregoing, and of Newton's second law, given the 
initial velocity in amount and direction, the starting-point, 
the initial amounts and directions of the acting forces and the 



74 



MECHANICS OF ENGINEERING. 



laws of their variation if they are not constant, we can resolve 
them into a single X and a single Y force at any instant, 
determine the X and Amotions independently, and afterwards 
the resultant motion. The resultant force is never in the direc- 
tion of the tangent to the path (except at a point of inflection). 
The relations which its amount and direction at any instant 
bear to the velocity, the rate of change of that velocity, and 
the radius of curvature of the path will appear in the next 
paragraph. 



y\i 



74. General Equations for the curvilinear motion of a ma- 
terial point in a plane. — The motion will be considered result- 
ing from the composition of 
independent X and Y motions, 
.Xand Y being perpendicular to 
each other. Fig. 79. In two 
consecutive equal times (each 




= dt), let dx and dx' = small 
spaces due to the X motion ; 
and dy and CK = dy', due to 
the Y motion. Then ds and 
ds' are two consecutive elements 
fig. 79. of the curvilinear motion. Pro- 

long ds, making BE = ds; then EE ' = d*x, CE= d*y, and 
CO = d*s {EO being perpendicular to BE). Also draw CL 
perpendicular to BG and call CL d 2 n. Call the velocity of 
the X motion v x , its acceleration p x ; those of the Y motion^ 
V y and p y . Then, 



v„ = 



d*x 
df 



A dV V 



dx dy dv, 

dt> Vy = ~di ; Px = ~dt 
For the velocity along the curve (i.e., tangent) 
v = ds -— dt, we shall have, since ds* = dx 2 -f- dy*, 
, (dsV fdxV . (dy\ a 

*=U = [dt) + KdiJ = v *+ v *' 



df* 



CO 



Hence v is the diagonal formed on v x and v v (as in § 71). 
Let j^ = the acceleration of v, i.e., the tangential acceleration^. 



CURVILINEAR MOTION OF A MATERIAL POINT. 75 

then^? e = d 2 s -— df, and, since d*s = the sum of the projec- 
tions of EF 'and CF on BC, i.e., d*s = d*x cos a-\- d*y sin a, 
we have 

tZ 2 5 d*x d*y 

df = df C0S a ~^ df sin a; Le *' -^ = P* C0S a +^ sin *" ( 2 ) 

By Normal Acceleration we mean the rate of change of the 
velocity in the direction of the normal. In describing the ele- 
ment AB = ds, no progress has been made in the direction of 
the normal BMi.e., there is no velocity in the direction of the 
normal; but in describing BC (on account of the new direc- 
tion of path) the point has progressed a distance CL (call it 
d 2 ri) in the direction of the old normal BH (though none in. 
that of the new normal CI). Hence, just as the tang. ace. 

ds' -- ds d*s . . . CL — zero d*n 

-™, so the normal acceJ. = 



df ~ dr ~~ df - df 

It now remains to express this normal acceleration (=j? n ) in 
terms of the X and Y accelerations. From the figure, CL 
= CM- ML, i.e., 

d'n = d*y cos a — d*x sin a {since EF = d*x] ; 

d?n d*y d*x . 

•■• df = if cos a ~ W 81U a - 

Hence L>n—L>y cos a — j^sin a (3) 

The norm. ace. may also be expressed in terms of the tang, 
velocity v, and the radius of curvature r, as follows : 
ds' = rda, or da = ds r ~ r ; also d*n = ds'doc, = ds n -f- r y 
. tfn (dsVl v* 

^-df^Kdt) P ° r ^=r W 

If now, Fig. 80, we resolve the forces X = Mp x and Y 
— Mp y , which at this instant account for the; 



< 



I r X and Y accelerations (M = mass of the 



a; 



material point), into components along the 
- \a \ tangent and normal to the curved path, we 
>P^M\ ,.--'' shall have, as their equivalent, a tangential 
force 



«o K ., 



Fig. 80. 



T = Mp x cos a -f- Mp y sin a, 



76 MECHANICS OF ENGINEERING. 

and a normal force 

iV" = Mp y cos a — Mp x sin a. 
But [see equations (2), (3), and (4)] we may also write 

T=Mp t = M^ t ; and W = Mp n = Mp . (5) 

Hence, if a free material point is moving in a curved path, 
the sum of the tangential components of the acting forces must 
equal (the mass) X tang, accel.; that of the normal components, 
= (the mass) X normal accel. = (mass) X (square of veloc. in 
path) -f- (rad. curv.). 

It is evident, therefore, that the resultant force (= diagonal 
on T and N or on Xand Y, Fig. 80) does not act along the tan- 
gent at any point, but toward the concave side of the path ; un- 
less r = go. 

JZadius of curvature. — From the line above eq. (4) we 
have d?n = ds* — r ; hence (line above eq. (3) ), ds 2 — r = 
d*y cos a — d*x sin a ; but cos a=dx-^-ds, and sin a=dy-i-ds, 

ds* , Q dx 7Q dy ds 3 , Tdxd^y — dyd?x~] 

■••-T"*VS-*VS or - = M\ _ M y j; 

i.e., — = dx*d\ -4- = dx*d (tan a), 

•'* r ~ \dt) T L\2*7 " <fc J ; 
<# tan afl 

'^Hy- s-J ( fi ) 

which is equally true if, for v x and tan a, we put v y and 

tan (90° —a), respectively. 

Change in the velocity square. — Since the tangential accelera- 

dv . _ d v _ 

tion -j-. =Pt> we nave ds-.- =p t ds; i.e., 

-j-dv =p t ds, or -y<#y =p t ds and .\ — - — =. I p t ds. (7) 

having integrated between any initial point of the curve where 
v = c, and any other point where v = v. This is nothing 
more than equation (III.), of § 50. 




CURVILINEAR MOTION OF A MATERIAL POINT. 77 

75. Normal Acceleration. Second Method. — Fig. 81. Let 
C be the centre of curvature and OD = 2r. Let OB' be a. 
portion of the oscillatory parabola (vertex at D \ N 
; any osculatory curve will serve). When \ 
ds is described, the distance passed over in 
the direction of the normal is AB ; for 2ds, 
it would be A ' B' = 4:AB (i.e., as the 
square of OB'; property of a parabola), 
and so on. Hence the motion along the normal is uniformly 
accelerated with initial velocity == 0, since the distance AB, 
varies as the square of the time (considering the motion along 
the curve of uniform velocity, so that the distance OB is di- 
rectly as the time). If p n denote the accel. of this uniformly 
accelerated motion, its initial velocity being = 0, we have (eq. 
2, § 56) AB = ip n dt\ i.e.,p n = 2JLB -r- df. __But from the 
similar triangles ODB and OAB we have, AB :ds::ds: 2r 9 
hence 2AB = ds* -f- r, .*. p n = ds* -i- rdtf = v 2 -^ r. 

76. Uniform Circular Motion. Centripetal Force. — The ve- 
locity being constant, jp f must be = 0, and .*. ^'(or^T'if there 
are several forces) must = 0. The resultant of all the forces, 
therefore, must be a normal force = (Mc* ~ r) = a con- 
stant (eq. 5, § 74). This is called the " deviating force," 
or "centripetal force ;" without it the body would continue 
in a straight line. Since forces always occur in pairs (§ 3), 
a "centrifugal force," equal and opposite to the "centri- 
petal" (one being the reaction of the other), will be found 
among the forces acting on the body to whose constraint the 
deviation of the first body from its natural straight course is 
due. For example, the attraction of the earth on the moon 
acts as a centripetal or deviating force on the latter, while the 
equal and opposite force acting on the earth may be called 

Ck the centrifugal. If a small block moving on a 

smooth horizontal table is gradually turned from 
its straight course AB by a fixed circular guide, 
^v_3^m tangent to AB at B, the pressure of the guide 
against the block is the centripetal force Me*-*- r 
directed toward the centre of curvature, while 




78 



MECHANICS OF ENGINEERING. 



///////■////'////■ 




the centrifugal force M& -f- r is the pressure of the block 
against the guide, directed away from that centre. The cen- 
trifugal force, then, is never found among the forces acting on 
the body whose circular motion we are dealing with. 

The Conical Pendulum, or governor- ball. — Fig. 82. If a 
material point of mass = M = G -r- g, suspended on a cord of 
length = I, is to maintain a uniform cir- 
cular motion in a horizontal plane, with a 
given radius r, under the action of gravity 
and the cord, required the velocity c to be 
given it. At B we have the body free. 
The only forces acting are G and the cord- 
fig. 83. tension P. The sum of their normal com- 

ponents, i.e., 2JV, must = Mc* -~ r, i.e., P sin a = Mc* -f- r ; 
but, since 2 (vert, comps.) = 0, P cos a = G. Hence 
G tan a = Gd 1 -^- gr\ .*. c = Vgr tan a. Let u = number of 
revolutions per unit of time, then u = e -i- %7tr = Vg — 2?r Vh; 
i.e., is inversely proportional to the (vertical projection)* of 
the cord-length. The time of one revolution is = 1 -f- u. 

Elevation of the outer rail on railroad curves (considera- 
tions of traction disregarded). — Consider a single car as a 
material point, and free, having a given 
velocity = c. P is the rail-pressure j^v f 

against the wheels. So long as the car £ -r— 

follows the track the resultant R of P \ 
.and G must point toward the centre of in- 
curvature and have a value = M& -f- r. 
But P= G tan or, whence tan a = tf-r- gr. 
If therefore the ties are placed at this 
angle a with the horizontal, the pressure 
will come upon the tread and not on the flanges of the wheels ; 
in other words, the car will not leave the track. (This is really 
the same problem as the preceding,) 

Apparent weight of a body at the equator. — This is less than 
the true weight or attraction of the earth, on account of the 
uniform circular motion of the body with the earth in its 
diurnal rotation. If the body hangs from a spring-balance, 




Fig. 84. 



CURVILINEAR MOTION OF A MATERIAL POINT. 79 

whose indication is G lbs. (apparent weight), while the true 
attraction is G' lbs., we have G' — G = M& -=- r. For M 
we may use G -^ g (apparent values); for r about 20,000,000 
ft.; for c, 25,000 miles in 24 hrs., reduced to feet per second. 
It results from this that G is < G' by -^k^G' nearly, and 
(since 17 a = 289) hence if the earth revolved on its axis seven- 
teen times as fast as at present, G would = 0, i.e., bodies 
would apparently have no weight, the earth's attraction on 
them being just equal to the necessary centripetal or deviating 
force necessary to keep the body in its orbit. 

Centripetal force at any latitude. — If the earth were a ho- 
mogeneous liquid, and at rest, its form would be spherical ; but 
when revolving uniformly about the polar diameter, its form 
of relative equilibrium (i.e., no motion of the particles relatively 
to each other) is nearly ellipsoidal, the polar diameter being an 
axis of symmetry. 

Lines of attraction on bodies at its surface do not intersect 
in a common point, and the centripetal force requisite to keep 
a suspended body in its orbit (a small circle of the ellipsoid), 
at any latitude /3 is the resultant of the attraction or true 
weight G' directed (nearly) toward the centre, and of G the 
tension of the string. Fig. 85. G is the apparent weight, in- 
dicated by a spring-balance and MA is its 
line of action (plumb-line) normal to the 
ocean surface. Evidently the apparent 
weight, and consequently g, are less than 
the true values, since N must be perpen- 
dicular to the polar axis, while the true 
values themselves, varying inversely as fkTST 

the square of MC, decrease toward the equator, hence the ap- 
parent values decrease still more rapidly as the latitude dimin- 
ishes. The following equation gives the apparent g for any 
latitude /?, very nearly (units, foot and second): 

g = 32.1808 - 0.0821 cos 2/3. 

(The value 32.2 is accurate enough for practical purposes.) 
&ince the earth's axis is really not at rest, but moving about 




80 



MECHANICS OF ENGINEERING. 



the sun, and also about the centre of gravity of the moon and 
earth, the form of the ocean surface is periodically varied, i.e., 
the phenomena of the tides are produced. 

77. Cycloidal Pendulum. — This consists of a material point 
at the extremity of an imponderable, flexible, and inextensible 
cord of length == Z, confined to the arc of a cycloid in a ver- 
tical plane by the cycloidal evolutes shown in Fig. 86. Let 
the oscillation begin (from rest) at A, a height = h above 





the vertex. On reaching any lower point, as B (height = z 
above 0), the point has acquired some velocity v, which is at 
this instant increasing at some rate = p t . Now consider the 
point free, Fig. 87; the forces acting are P the cord-tension, 
normal to path, and G the weight, at an angle <p with the 
path. From § 74, eq. (5), 2T = Mp t gives 

G cos <p + P cos 90° = (G + g)p t ; .\p t = g cos <p 

Hence (eq. (7), § 74), vdv = p t ds gives 

vdv = g cos cpds ; but ds cos cp = — dz ; .*. vdv = — gdz. 

Summing between A and B, we have 

1—0 

the same as if it had fallen freely from rest through the height 
h — z. {This result evidently applies to any form of path 
when, besides the weight G, there is but one other force, and 
that always normal to the path.) 

From XZT = Mv* ~ n, we have P — G sin cp = Mv* -r-r lr 



CURVILINEAR MOTION OP A MATERIAL POINT. 81 

whence _P, the cord-tension at any point, may be found (here 
r x — the radius of curvature at any point = length of straight 
portion of the cord). 

To find the time of passing from A to 0, & half-oscillation, 
substitute the above value of v* in v = ds -s- dt, putting ds % 
= dx* + dz*, and we have dt = (dx* -f- dz*) -r- [2g(h — zj]* 
To find dx in terms of dz, differentiate the equation of the 
curve, which in this position is 



x = r ver. sin. -1 (z -=- r) -f- V2rz — z* ; 
whence 

r/Iz (r — z)dz _ (2r — z)dz 

' V2rz — z* V2rz - z* ~ VZrz — z*' 

.-.dx* = [t- 1 ]^ 
(r = radius of the generating circle). Substituting, we have 



dt 



Vr 



Yhz 



— » J 



Hence the whole oscillation occupies a time = n Yl -f- g 
(since I = 4r). This is independent of A, i.e., the oscillations 
are isochronal. This might have been proved by showing that 
pt is proportional to OB measured along the curve ; i.e., that 
the motion is harmonic. (§ 59, Prob. 2.) 

78. Simple Circular Pendulum. — If the material point oscil- 
lates in the arc of a circle, Fig. 88, proceeding 
as in the preceding problem, we have finally, 
after integration by series, as the time of a full 
oscillation, 

©r°* An 1 h . 9 v i i 



Hence for a small h the time is nearly n Yl -=- g, and the os- 



82 



MECHANICS OF ENGINEERING. 



cillations nearly isochronal. (For the Compound Pendulum, 
see § 117.) 

79. Change in the Velocity Square. — From eq. (7), § 74, we 
have ^('y 2 — c 2 ) =fp t ds. But, from similar triangles, du be- 
ing the projection of any ds of the path upon the resultant 
force R at that instant, Rdu = Tds (or, Prin. of Virt. Yels. 
§ 62, Rdu = Tds + i^X 0). T and iTare the tangential and 
normal components of R. Fig. 89. Hence, finally, 

iMv*-iMe'=/Rdu, (a) 

for all elements of the curve between any two points. In gen- 



<_(Z-S— v 



*T — * 



Nt 



. ^1R 

Fig. 89. 



eneral R is different in amount and direc- 
tion for each ds of the path, but du is the 
distance through which R acts, in its own 
direction, while the body describes any ds ; 
Rdu is called the work done by R when ds is described by the 
body. The above equation is read : The difference between the 
initial and final kinetic energy of a body = the work done by 
the resultant force in that portion of the path. 

(These phrases will be further spoken of in Chap. YI.) 
Application of equation (a) to a planet in its orbit about 
the sun. — Fig. 90. Here the only force at any instant is the at- 
traction of the sun R = C -=- u? (see Prob. 3, § 59), 
where G is a constant and u the variable radius 
vector. As u diminishes, v increases, therefore 
dv and du have contrary signs ; hence equation > 
(a) gives (c being the velocity at some initial ° 
point 0) 



^Mv?--Mc\ 



pdu = n__^ (h) 

Ju u \_u x u J 7 



aA+ 



26fl_ 



U, 



a 



which is independ- 




ent of the direction of the initial velocity c. 

Note. — If u were = infinity, the last member of equation (b) would re- 
duce to G -5- tii, and is numerically the quantity called potential in the 
theory of electricity. 



CURVILINEAR MOTION OF A MATERIAL POINT. 83 



Application of eq. (a) to a projectile in vacuo. 
body's weight, is the only force acting, and O 
therefore = B, while M— G -f- g. There- 
fore equation (a) gives 

G < 

9 ' 



G, the 



-=GJ dy=G yi ; 





,-. v l = vV + 2gy\, which is independent of FlG . 91. 

the angle, a, of projection. 

Application of equation (a) to a body sliding, without fric- 
tion, on a fixed curved guide in a vertical plane ; initial velo- 
city = c at 0. — Since there is some pressure at each point be- 
tween the body and the guide, to consider the body free in 
space, we must consider the guide removed and that the body 

describes the given curve as a re- 
sult of the action of the two forces, 
its weight G, and the pressure P, 
of the guide against the body. G 
is constant, while P varies from 
point to point, though always (since 
there is no friction) normal to curve. 
At any point, R being the resultant 
of G and P, project ds upon P, thus obtaining du ; on G, 
thus obtaining dy ; on P, thus obtaining zero. But by the 
principle of virtual velocities (see § 62) we have Pdu = Gdy 
-fPx zero = Gdy, which substituted in eq. (a) gives 

j\(< - <?) =£ 1 Gdy=Gf"dy=Gy l ; .-.*,,= V^fW, 

and therefore depends only on the vertical distance fallen 
through and the initial velocity, i.e., is independent of the 
form of the guide. 

As to the value of P, the mutual pressure between the guide 
and body at any point, since '2JN' must equal Mv 1 -f- r,r being 
the variable radius of curvature, we have, as in § 77, 

P — G sin <p = Mv' -r^; .. P = G[sin <p+( v 7 -T- gr)]. 

As, in general, cp and r are different from point to point of 



84 MECHANICS OF ENGINEERING. 

the path, P is not constant. (The student will explain how P 
may be negative on parts of the curve, and the meaning of 
this circumstance.) 

80. Projectiles in Vacuo. — A ball is projected into the air 
|Y (whose resistance is neglected, hence the 

""7° ./"""TF phrase in vacuo) at an angle = a with the 
/J( G \1 horizontal ; required its path ; assuming it 
\ \ I _. confined to a vertical plane. Resolve the 

l o • i Cx_ J£ a ... 

IZ^r-i.. ■ "> motion into independent horizontal (X) 
fig. 93. an d vertical (Y) motions, G> the weight, 

the only force acting, being correspondingly replaced by its 
horizontal component = zero, and its vertical component 
= — G. Similarly the initial velocity along X= c x = c cos <*„, 
along Y, = c y = csin a . The JT acceleration =p x = -r- M 
= 0, i.e., the X motion is uniform, the velocity v x remains 
= c x = c cos a at all points, hence, reckoning the time from 0, 
at the end of any time t we have 

x = c(cos a )t (1) 

In the Y motion, p y = (— G) -f- M = — g, i.e., it is uniformly 
retarded, the initial velocity being c y = c sin a ; hence, after 
any time t, the ^velocity will be (see § 56) v v = c sin a — gt y 
while the distance 

y = c(sin a Q )t - \gt (2) 

Between (1) and (2) we ma} 7 eliminate t, and obtain as the 
equation of the trajectory or path 

gx* 
y = x tan a. — — -^ — 5 — . 
y ° 2c 2 cos 2 a 

For brevity put c* = 2gh, h being the ideal height due to the 
velocity c, i.e., c 2 -7- 2g (see § 53 ; if the ball were directed ver- 
tically upward, a height h = c* -r- 2<? would be actually at- 
tained, ar being = 90°), and we have 

a? 2 

y = x tan a, — --7 3 (3) 

* ° 4A cos 2 « x 7 

This is easily shown to be the equation of a parabola, with its 
axis vertical. 



CURVILINEAR MOTION OF A MATERIAL POINT. 85 

The horizontal range. — Fig. 94. Putting y = in equa- 
tion (3), we obtain 

x tan a. — T1 5 — = 0, Y i o 

[_ ° 4A cos 2 a J 

which is satisfied both by x = (i.e., at the 

origin), and by x = 4:h cos cx sin a Q . Hence * 

the horizontal range for a given c and a Q is FlG - 94 - 

x r = 4A cos <*„ sin ar = 2A sin 2<* . 

For a = 45° this is a maximum (c remaining the same), 
being then = 2A. Also, since sin 2a = sin (180° — 2a ) = 
sin 2(90° — ar ), therefore any two complementary angles of 
projection give the same horizontal range. 

Greatest height of ascent / that is, the value of y maximum, 
■= y m - — Fig. 94. Differentiate (3), obtaining 

dy x 

dx " ° 2A cos 2 aj 

which, put = 0, gives x = 2A sin a cos ar , and this value of 
a? in (3) gives y m = h sin 2 or . 

(Let the student obtain this more simply by considering the 
T motion separately.) 

To strike a given point ; c being given and a required. — 
Let x' and y' be the co-ordinates of the given point, and a/ 
the unknown angle of projection. Substitute these in equa- 
tion (3), h being known = & -f- 2g, and we have 

x n 1 

y' = x' tan a\ — -n i — i- Put cos 2 or/ = . - — = — 7 

* ° 4A cos 2 or/ ■ 1 + tan 2 <z " 

and solve for tan a \ whence 

tan a' = [2 A ± ^4A 2 - x"> - 4Ay'] -=- a/. . . (4) 

Evidently, if the quantity under the radical in (4) is negative, 
tan a ' is imaginary, i.e., the given point is out of range with 
the given velocity of projection c = V~2gh ; if positive, tan a/ 
has two values, i.e., two trajectories may be passed through 
the point ; while if it is zero, tan a ' has but one value. 

The envelope, for all possible trajectories having the same 



86 



MECHANICS OF ENGINEERING. 









0" £ 








/ 




// 







\ \ 



Fig. 95. 



initial velocity c (and hence the same h) ; i.e., the curve tan- 
gent to them all, has but one point of contact with any one of 
them ; hence each point of the envelope, Fig. 95, must have 

co-ordinates satisfying the con- 
dition, 4A 2 — x n — 4:hy' = ; i.e. 
(see equation (4) ), that there is 
but one trajectory belonging to it. 
Hence, dropping primes, the 
equation of the envelope is 4A 2 — 
x 2 — ±hy = 0. Now take 0" as a 
new origin, a new horizontal axis X", and reckon y" positive 
downwards ; i.e., substitute x = x" and y = h — y" . The 
equation now becomes x"* = 4hy f/ ; evidently the equation of 
a parabola whose axis is vertical, whose vertex is at 0", and 
whose parameter = 4A = double the maximum horizontal 
range. is therefore its focus. 

The range on an inclined plane. — Fig. 96. Let OG be 
the trace of the inclined plane ; its equation y 
is y = x tan /?, which, combined with the 
equation of the trajectory (eq. 3), will give 

the co-ordinates of their intersection G. C C......:^...:. ; — ^ 

That is, substitute y = x tan /? in (3) and fig. 96. 

solve for x, which will be the abscissa x„ of G. This gives 




tan a. — tan /3 = 



sin a sin f3 sin (a — /?) 
cos <x cos ft cos a cos /3 ' 



4A cos 2 a 

.*. x x = 4Acos a sin (ar — fi) -r- cos /?, and the range OG r 
which = x x -r- cos /?, is = (4A -f- cos 2 /3) cos <* sin (a — /?). (5) 

TA<? maximum range on a given inclined plane, /?, c (and 
.*. h), remaining constant, while a varies. — That is, required 
the value of a which renders OG a maximum. Differentiat- 
ing (5) with respect to ar , putting this derivative — 0, we have 
[4A ~ cos 2 /?] [cos a cos (a — /?) — sin a sin (a — /?)] = ; 
whence cos \_a + (*, - /?)] = 0; i.e., 2^ - fi = 90°; or, 
ar = 45° -f- ^/?, for a maximum range. By substitution this 
maximum becomes known. 

The velocity at any point of the path is v = VvJ + v y * = 



CURVILINEAR MOTION OF A MATERIAL POINT. 



87 



y& — 2ctg sin a -\- g*tf (see the first part of this § 80) ; while 
the time of passage from to any point whose abscissa is x is 
t = x -r- c cos a ; obtained from equation (1). E.g., to reach 
the point B, Fig. 94, we put x = x r = 4:h sin a cos ar, and ob- 
tain t r = 2c sin a -f- g. This will give the velocity at B = 

V? = 0. 



81. Actual Path of Projectiles. — Small jets of water, so long as 
they remain unbroken, give close approximations to parabolic 
paths, as also any small dense object, e.g., a ball of metal, hav- 
ing a moderate initial velocitv. The course of a cannon-ball, 
however, with a velocity of 1200 to 1400 feet per second is 
much affected by the resistance of the air, the descending 
branch of the curve being much steeper than the ascending; 
see Fig. 96a. The equation of this curve has not yet been 
determined, but only the expression for the slope (i.e., 
dy : dx) at any point. See Professor Bart- . 

lett's Mechanics, § 151 (in which the body 
is a sphere having no motion of rotation). 
Swift rotation about an axis, as well as an 
unsymmetrical form with reference to the 
direction of motion, alters the trajectory 
still further, and may deviate it from a vertical plane, 
presence of wind would occasion increased irregularity 
Johnson's Encyclopaedia, article " Gunnery." 




Fig. 96a. 



The 
See 



82. Special Problem (imaginary ; from "Weisbach's Mechan- 
ics. The equations are not homogeneous). — Suppose a ma- 
terial point, mass = M, to start 
from the point O, Fig. 97, with 
a velocity = 9 feet per second 
along the — Y axis, being sub- 
jected thereafter to a constant 
attractive X force, of a value X 



= 12M, and to a variable Y 

force increasing with the time 

fig. 97. (in seconds, reckoned from 0\ 




88 MECHANICS OF ENGINEERING. 

viz., Y= SMi. Required the path, etc. For the X motion 
we have^ = JT-f- M — 12, and hence 

dv x = I p x dt = 12 / dt; i.e., v x == 12t; 

dx= v x dt ; i.e., x = 12 / ^ = 6f . . (1) 

dv y =S / tdt; 

9 ^o 

i.e.,v y -\-9 = 4tf, and I dy = / v y dt; 

.-. y = ±f fdt - <df dt, or y = %f - 9*. . . (2) 

Eliminate t between (1) and (2), and we have, as the equa- 
tion of the "path, 

» = 4(fMf)' m 

which indicates a curve of the third order. 
The velocity at any point is (see § 74, eq, (1) ) 



*> = 4V + <=4* , + 9 (4) 

TAtf length of curve measured from (9 will be (since v = 
ds -7- ^) 

5 = /*<& =y vdt = ±f fdt + 9 J* dt = p 3 + 9t. (5) 

The slope, tan «, at any point = v y ~- v x = (4tf — 9) ~ 12t, 
dtzna 4* 2 + 9 

and ■'■ ST = 1ST (6) 

7%0 radius of curvature at any point (§ 74, eq. (6) ), sub- 
stituting v x — 122, also from (4) and (6), is 

, = ^[,/-^]=> +9 r, • • <*> 

and the normal acceleration —v^-^-r (eq. (4), § 74), becomes 
from (4) and (7) p n = 12 (ft. per square second), a constant. 
Hence the centripetal or deviating force at any point, i.e., the 



CURVILINEAR MOTION OF A MATERIAL POINT. 89 

2 IT of the forces X and Y, is the same at all points, and = 
Mtf + r = 12Jf. 

From equation (3) it is evident that the curve is symmetrical 
-about the axis X. Negative values of t and s would apply to 
points on the dotted portion in Fig. 97, since the body may be 
considered as having started at any point whatever, so long as 
all the variables have their proper values for that point. 

(Let the student determine how the conditions of this motion 
could be approximated to experimentally.) 

83. Relative and Absolute Velocities. — Fig. 98. Let Mbe& 
material point having a uniform motion of velocity v a along a 
straight groove cut in the deck of a steamer, which itself has 
a uniform motion of translation, of velocity v v over the bed of 
3, river. In one second M ad- \ / 

vances a distance % along the 
groove, which simultaneously has /^ /"' m 



iv. 




moved a distance v. — AB with I \ . /// V V j> ^> 

the vessel. The absolute path of ^^'-'""^!zH^^Z *"" 

M during the second is evidently fig. 98. 

w (the diagonal formed on v x and v 2 ), which may therefore be 
called the absolute velocity of the body (considering the bed 
of the river as fixed) ; while v 2 is its relative velocity, i.e., rela- 
tive to the vessel. If the motion of the vessel is not one of 
translation, the construction still holds good for an instant of 
time, but v l is then the velocity of that point of the deck over 
which M is passing at this instant, and v 2 is M's velocity rela- 
tively to that point alone. 

Conversely, if M be moving over the deck with a given 
absolute velocity = to, v x being that of the vessel, the relative 
velocity % may be found by resolving w into two components, 
one of which shall be v 1 ; the other will be v t . 

If w is the absolute velocity and direction of the wind, the 
vane on the mast-head will be parallel to MT, i.e., to v 9 the 
relative velocity ; while if the vessel be rolling and the mast- 
head therefore describing a sinuous path, the direction of the, 
vane varies periodically. 



90 MECHANICS OF ENGINEERING. 

Evidently the effect of the wind on the sails, if any, will 
depend on v a the relative, and not directly on w the absolute, 
velocity. Similarly, if w is the velocity of a jet of water, and 
v l that of a water-wheel channel, which the water is to enter 
without sudden deviation, or impact, the channel-partition 
should be made tangent to v 2 and not to w. 

Again, the aberration of light of the stars depends on the 
same construction ; v l is the absolute velocity of a locality of the 
earth's surface (being practically equal to that of the centre) ; 
w is the absolute direction and velocity of the light from a 
certain star. To see the star, a telescope must be directed 
along MT, i.e., parallel to v 2 the relative velocity; just as in 
the case of the moving vessel, the groove must have the direc- 
tion MT. if the moving material point, having an absolute 
velocity w, is to pass down the groove without touching its 
sides. Since the velocity of light = 192,000 miles per second 
= w, and that of the earth in its orbit = 19 miles per second 
= ^j, the angle of aberration SMT, Fig. 98, will not exceed 
20 seconds of arc ; while it is zero when w and v 1 are parallel. 

Returning to the wind and sail-boat, it will be seen from 
Fig. 98 that when v 1 — or even > w, it is still possible for v 2 
to be of such an amount and direction as to give, on a sail 
properly placed, a small wind-pressure, having a small fore-and 
aft component, which in the case of an ice-boat may exceed 
the small fore-and-aft resistance of such a craft, and thus v x will 
be still further increased ; i.e., an ice-boat may sometimes travel 
faster than the wind which drives it. This has often been 
proved experimentally on the Hudson River. 



MOMENT OF INEKTIA. 91 



CHAPTER TV. 

MOMENT OF INERTIA. 

[Note. — For the propriety of this term and its use in Mechanics, see 
§114; for the present we are only concerned with its geometrical nature.] 

85. Plane Figures. — Just as in dealing with the centre of 
gravity of a plane figure (§ 23), we had occasion to sum the 
series fzdF, z being the distance of any element of area, dF, 
from an axis ; so in subsequent chapters it will be necessary to 
know the value of the series fz*dF for plane figures of various 
shapes referred to various axes. This summation fz*dF of 
the products arising from multiplying each elementary area of 
the figure by the square of its distance from an axis is called 
the moment of inertia of the plane figure with respect to the 
axis in question • its symbol will be I. If the axis is perpen- 
dicular to the plane of the figure, it may be named the polar 
mom. of inertia (§ 94) ; if the axis lies in the plane, the rec- 
tangular mom. of inertia (§§ 90-93). Since the / of a plane 
figure evidently consists of four dimensions of length, it may 
always be resolved into two factors, thus /= FA?, in which 
F = total area of the figure, while Jc = VI -f- F, is called the 
radius of gyration, because if all the elements of area were 
situated at the same radial distance, k, from the axis, the 
moment of inertia would still be the same, viz., 

I = JVdF = kfdF = Fk\ 

86. Rigid Bodies. — Similarly, in dealing with the rotary 
motion of a rigid body, we shall need the sum of the series 
fp 2 dM, meaning the summation of the products arising from 
multiplying the mass dM of each elementary volume dVof a 



92 MECHANICS OF ENGINEERING. 

rigid body by the square of its distance from a specified axis. 
This will be called the moment of inertia of the body with 
respect to the particular axis mentioned (often indicated by a 
subscript), and will be denoted by I. As before, it can often 
be conveniently written M¥, in which M is the whole mass, 
and k its "radius of gyration" for the axis used, k being 
= Vl-r- M. If the body is homogeneous, the heaviness, y, of 
all its particles will be the same, and we may write 

I=fp>dM= (y - g)fp'dV= ( r - g) VV. 

87. If the body is a homogeneous plate of an infinitely small 
thickness = r, and of area = F, we have I — (y ^ g)fp*d V 
= (y ~- g)rfp*dF\ i.e., = (y -f- g) X thickness X mom. iner- 
tia of the plane figure. 




Two Parallel Axes. Reduction Formula. — Fig. 99. Let 
Z and Z be two parallel axes. Then I z 
=fp*dM, and l z> =fp n dM. But d being 

dVNg, the distance between the axes, so that a? 

—¥ + tf) + d * - 2aa> - 2by, and .-. 
I z , =fp 2 dM+dfdM- 2afxdM 

-2bfydM. . (1) 

But fp'dM = I z ,fdM= M, and from the 
theory of the centre of gravity (see § 23, eq. (1), knowing that 
dM = yd V+ g, and .\ that [fyd V] -4- g= M ) we have fxdM 
= Mx di.ud.fydM = My ; hence (1) becomes 

l z , = I z +M(d*-2ax-2by), . ... (2) 

in which a and b are the x and y of the axis Z\ x and y refer 
to the centre of gravity of the body. If Z is a gravity-axis 
(call it g), both x and y = 0, and (2) becomes 

I z , =I g + Md* or k z ,* = V + d\ . . (3) 

It is therefore evident that the mom. of inertia about a grav- 
ity-axis is smaller than about any other parallel axis. 

Eq. (3) includes the particular case of a plane figure, by 



MOMENT OF INERTIA. 



93 



•writing area instead of mass, i.e., when Z (now g) is a gravity- 
axis, 

(4) 



I z ,=I g + FcT. 



89. Other Reduction Formulae ; for Plane Figures. — (The axes 
here mentioned lie in the plane of the figure.) For two sets 
of rectangular axes, having the same origin, the following holds 
good. Fig. 100. Since 

I x =ffdF, and I Y =fx 2 dF, 

we have I x + I Y =f(x 2 + f)dF 

Similarly, , J n -f I v =f{tf + u 2 )dF 

But since the x and y of any dFh&ve the same hypothenuse as 
the u and v, we have v* + u* = x*-\- y*; .'. T x -\- 1 Y = I v -\-I V: >. 





Fig. 100. 



Fig. 100a. 



Let Xbe an axis of symmetry ; then, given I x and I Y (O is 
anywhere on X) % required I v , U being an axis through O and 
making any angle a with X. 

l v =fv*dF=f(y cos a — x sin otfdF\ i.e., 

I n = cos 2 cxfy'dF — 2 sin a cos afxydF-\- sin 3 afx*dF. 

But since the area is symmetrical about X, in summing up the 
products xydF, for every term a?( -f- 2/)<#i^, there is also a term 
a?( — y)dF to cancel it ; which gives fxydF = 0. Hence 

J n = cos 2 tfTx -f- sin 2 aI Y . 

The student may easily prove that if two distances a and b 
be set off from O on X and Y respectively, made inversely 
proportional to V I x and Vl Y , and an ellipse described on a and 
b as semi-axes ; then the moments of inertia of the figure about 



94 



MECHANICS OF ENGINEERING. 



any axes through are inversely proportional to the squares 
of the corresponding semi-diameters of this ellipse ; called 
therefore the Ellipse of Inertia. It follows therefore that the 
moments of inertia about all gravity-axes of a circle, or a 
regular polygon, are equal ; since their ellipse of inertia must 
be a circle. Even if the plane figure is not symmetrical, an 
" ellipse of inertia" can be located at any point, and has the 
properties already mentioned ; its axes are called the principal 
■axes for that point. 

90. The Rectangle. — First, about its base. Fig. 101. Since 
all points of a strip parallel to the base 
.£.._,. < ft_.» j iave ^ same co-ordinate, z, we may take 

dz the area of such a strip for dF= bdz; 
g .-. I B =f h z i dF= bf^z'dz 

l_0 



dz 






in 



Fig. 101. 



Fig. 102. 



Secondly, about a gravity-axis parallel to base. 

z*dz = ^bh\ 

Thirdly, about any other axis in its plane. Use the results 
already obtained in connection with the reduction-formulse of 
§§88, 89. 

90a. The Triangle. — First, about an axis through the vertex 

and parallel to the base ; i.e., I v <. & » ^ % ^ R 

in Fig. 103. Here the length j V y. ^ \hY~ 

of the strip is variable ; call it y. ^ 

From similar triangles \\ / \ / 

y = (b -+ h)z; 




Fig. 103. 



Fig. 104. 



.-. I v =.fz*dF= fz'ydz = ( I -r- h)f z*dz = \bh\ 

Secondly, about g, a gravity -axis parallel to the base. Fig. 
104. From § 88, eq. (4), we have, since F— %bh and 

d = \h, I g = l v - Fd? = W - \bh . V* 2 = ^bh\ 



MOMENT OF INERTIA. 



95 



Thirdly, Fig. 104, about the base ; I B = % From § 88, eq. 
(4), I B = I g + Fd% with d — \h ; hence 

I B = ^bh* + %bh . |A 2 = ^bh\ 

91. The Circle. — About any diameter, as g, Fig. 105. Polar 
co-ordinates, I g = fz'dF. Here we take dF= area of an ele- 
mentary rectangle = pdcp . dp, while z = p sin <p. 




t A 

1 E 


: f H 




r*5c 


—at — 


:tbi 


it 


4 

— i 


■-ft 



Fig. 105. 



Fig. 106. 



l g — J I \p sin cpfpdcpdp —J sin 8 cpdcpj p*dp\ 

= — / sin 2 ^9> = -j- / i(l — cos 2<p)d<p 
r* r 2n \~\ 1 1 



= 4- 



\2^~4 sm2 ?7 



J 2 *_ )-(0-0)]. Le.,I g = \*r\ 



92. Compound Plane Figures. — Since I = fz' i dF is an in- 
finite series, it may be considered as made up of separate 
groups or subordinate series, combined by algebraic addition, 
corresponding to the subdivision of the compound figure into 
component figures, each subordinate series being the moment of 
inertia of one of these component figures; but these separate 
moments must all be referred to the same axis. It is con- 
venient to remember that the {rectangular) 7 of a plane 
figure remains unchanged if we conceive some or all of its 
elements shifted any distance parallel to the axis of refer- 
ence. E.g., in Fig. 106, the sum of the I B of the rectangle OF, 
and that of FD is = to the I B of the imaginary rectangle 



MECHANICS OF ENGINEERING. 



formed by shifting one of them parallel to B, until it touches 
the other ; i.e., I B of CJS+ I B of FD = ^h* (§ 90). Hence 
the I B of the T shape in Fig. 106 will be = I B of rectangle 
AD - I B of rect. OF- I B of rect. FD. 

That is, I B of T = i[bh 3 - \K\ ■ • - (§ 90). . . (1) 

About the gravity-axis, g, Fig. 106. To find the distance d 
from the base to the centre of gravity, we may make use of 
eq. (3) of § 23, writing areas instead of volumes, or, experi- 
mentally, having cut the given shape out of sheet-metal or 
card-board, we may balance it on a knife-edge. Supposing d 
to be known by some such method, we have, from eq. (4) of 
§ 88, since the area F= bh — bji^ I g = I B — Fd 2 ; 

i.e., I g = i[bh? - &A 3 ] - Q>h - &AK- • • ( 2 ) 
The double-T (orx), and the box forms of Fig. 106#, if 

symmetrical about the gravity- 
axis <7, have moments of inertia 
alike in form. Here the grav- 
ity-axis (parallel to base) of the 
compound figure is also a grav- 
ity axis (parallel to base) of each 
of the two' component rectangles, of dimensions b and h, b x and 
A l5 respectively. 

Hence by algebraic addition we have (§ 90), for either com- 
pound figure, 

J,= *[»•■- W] (3) 

(If there is no axis of symmetry parallel to the base we must 
proceed as in dealing with the T-form.) Similarly for the ring,. 







-6- 








b 


.... -v 




1 












s 


... 


+ i 

i h 

fcf 




>-tb? 


-g- 


..yj. 


<\~b r * 


fp 
















—X— 



Fig. 106a. 





Fig. 10?. 



Fig. 108. 



Fig. 107, or space between two concentric circumferences, we 
have ? about any diameter or g (§ 91), 



J g = l«-r:). 



(±> 



MOMENT OF INERTIA. 97 

The rhombus about a gravity-axis, g, perpendicular to a 
diagonal, Fig. 108. — This axis divides the figure into two 
equal triangles, symmetrically placed, hence the I g of the 
rhombus equals double the moment of inertia of one triangle 
about its base ; hence (§ 90«) 

l g = 2 . ^(W - A^ 3 (5) 

(The result is the same, if either vertex, or both, be shifted 
any distance parallel to AB.) 

For practice, the student may derive results for the trapezoid / 
for the forms in Fig. 106, when the inner corners are rounded 
into equal quadrants of circles; for the double- T, when the 
lower flanges are shorter than the upper; for the regular 
polygons, etc. 

93. If the plane figure be bounded, wholly or partially, by 
curves, it may be subdivided into an infinite number of strips, 
and the moments of inertia of these (referred to the desired 
axis) added by integration, if the equations of the curves are 
known ; if not, Simpson's Rule, for a finite even number of 
strips, of equal width, may be employed for an approximate 
result. If these strips are parallel to the axis, the / of any one 
strip = its length X its width X square of distance from axis ; 
while if perpendicular to, and terminating in, the axis, its 
/ = J its width X cube of its length (see § 90). 

A graphic method of determining the moment of inertia of 
any irregular figure will be given in a subsequent chapter. 

94. Polar Moment of Inertia of Plane Figures (§ 85). — Since 
the axis is now perpendicular to the plane of the figure, inter- 
secting it in a point, 0, the distances of the ele- 
ments of area will all radiate from this point, 
and would better be denoted by p instead of z ; 
hence, Fig. 109,/pVZ.Fis the polar moment of / 
inertia of any plane figure about a specified 
point ; this may be denoted by I p . But p 2 Fig. 109. 
= a? 2 -j" y 2 , for each dF\ hence 

I, =A* + y'¥F=fx'dF+fy>dF= 1 Y + I x . 



98 MECHANICS OF ENGINEERING. 

i.e., the polar moment of inertia about any given point i/n 
the plane equals the sum of the rectangular moments of iner- 
tia about any two axes of the plane figure, which intersect at 
right angles in the given point. We have therefore for the 
circle about its centre 



I p — \itr* + \nr* = 
For a ring of radii r x and r 2 , 

I p = M^' - <) ; 
For the rectangle about its centre, 

I, = ^W + T \hV = &bh(b> + A') ; 
For the square, this reduces to 

(See §§90 and 91.) 

95. Slender, Prismatic, Homogeneous Rod. — Keturning to the 

moment of inertia of rigid bodies, or solids, we begin with that 
of a material line, as it might be called, about 
an axis through its extremity making some an- 
gle a with the rod. Let I = length of the rod, 
F its cross-section (very small, the result being 
strictly true only when F = 0). Subdivide 
fig. no. the rod into an infinite number of small prisms, 

each having i^as a base, and an altitude = ds. Let y = the 
heaviness of the material ; then the mass of an elementary 
prism, or dM, = (y -f- g)Fds, while its distance from the axis 
Z is f> = s sin a. Hence the moment of inertia of the rod 
with respect to Z as an axis is 

I z = J t p*dM= (y -f- g)F 'sin 3 afs'ds = i(y + g)Fl z sin a a. 

But yFl -f- g = mass of rod and I sin a = a, the distance of 
the further extremity from the axis ; hence I z = %Ma* and 
the radius of gyration, or h, is found by writing \Ma x =^ MJc* ; 
.-. F = \a\ or Jc = V\a (see § 86). If a = 90°, a = I. 

96. Thin Plates. Axis in the Plate. — Let the plates be homo- 
geneous and of small constant thickness = r. If the surface of 




MOMENT OF INERTIA. 99 

the plate be = F, and its heaviness y, then its mass = yFt -f- g. 

From § 87 we have for the plate, about any axis, 

I=z (y -r- g)t X mom. of inertia of the plane figure formed by 

the shape of the plate . . . . (1) 

Rectangular plate. Gravity -axis parallel to base. — Dimen- 
sions b and h. From eq. (1) and § 90 we have 

Similarly, if the base is the axis, I B = -J-3/A 2 , .*. ¥ = ^A 3 . 
Triangular plate. Axis through vertex parallel to base. — 
From eq. (1) and § 90a, dimensions being b and A, 

I v =(y-r- g)T\bh? = {yibhr ~ g)W = JJ/A 2 ; .-. & 2 = \h\ 

Circular plate, with any diameter as axis. — From eq. (1) 
and § 91 we have 

l* = (r -*■ <7)4^ 4 = (r^ - <7)i^ 2 = P^ 2 ; /. & 2 = &*• 

97. Plates or Right Prisms of any Thickness (or Altitude). 
Axis Perpendicular to Surface (or Base). — As before, the solid is 
homogeneous, i.e., of constant heaviness y; | z 

let the altitude = A. Consider an elementary ^tV 

prism, Fig. Ill, whose length is parallel to the J^r 9 * \ ^"^ 
axis of reference Z. Its altitude = A = that | T~ 

of the whole solid ; its base = dF = an element 7 
of i^the area of the base of solid ; and each * 
point of it has the same p. Hence we may Fig. in. 
take its mass, = yhdF-z- g, as the dM In summing the series 
l z =fp*dM; 
.'.I z =(yh+g)fp*dF 

= (yh -f- g) X polar mom. of inertia of base. . . (2) 

By the use of eq. (2) and the results in § 94 we obtain the 
following : 

Circular plate, or right circular cylinder, about the geo- 
metrical axis, r = radius, A = altitude. 

I* = 0^ -*- 9ft*r* = (yhnr 1 -s- gft? = \Mr\ .-. ¥ =%r\ 

Right par allelopiped or rectangular plate. — Fig. 112, 
I g = (yh - g^bbM + V) = J/ T Vf ; .'. ff = A^. 



100 



MECHANICS OF ENGINEERING. 



For a hollow cylinder^ about its geometric axis, 





Fig. 112. 



Fig. 113. 



98. Circular Wire. — Fig. 113 (perspective). Let Z be a 
gravity-axis perpendicular to the plane of the wire ; X and Y 
lie in this plane, intersecting at right angles in the centre 0» 
The wire is homogeneous and of constant (small) cross-section. 
Since, referred to Z, each d M has the same p — r, we have 
l z z=zf r ' i dM= Mr\ Now I x must equal 7 F , and (§ 94) their 
sum = Z z , 



.-. I x , or Jx 



•Xj 



IMr* 



and 



Jc x , or by = i^ 2 . 




99. Homogeneous Solid Cylinder, about a diameter of its base, 
— Fig. 114. I x = ? Divide the cylinder into an infinite num- 
ber of laminae, or thin plates, parallel to the 
base. Each is some distance z from X, of 
thickness dz, and of radius r (constant). In 
each draw a gravity-axis (of its own) parallel to 

fig. 114. X. We may now obtain the I x of the whole 

cylinder by adding the I x s of all the laminae. The I g of any- 
one lamina (§96, circular plate) = its mass X i^ 2 ; hence its 
I x (eq. (3), § 88) = its I g -\- (its mass) X s 2 . Hence for the 
whole cylinder 

f * = £ &&*** + 9)W + O] 

= (*r*y -s- g\y£ dz +f\*dz~\-, 

i.e., I x = (xr'hy - g)(tf + W) = Mi^ + W). 

100. Let the student prove (1) that if Fig. 114 represent 
any right prism, and h F denote the radius of gyration of any 
one lamina, referred to its gravity-axis parallel to X, then the 
I x of whole prism = M(k F * -\- -JA 2 ); and (2) that the moment 



MOMENT OF INERTIA. 



101 




of inertia of the cylinder about a gravity-axis parallel to the 
base is = M(fr + T \h'). 

101. Homogeneous Right Cone. — Fig. 115. First, about an 
axis V, through the vertex and parallel to the base. As before, 
divide into laminae parallel to the base. Each is a 
circular thin plate, but its radius, x, is not = r, but, 
from proportion, is x = (r -j- h)z. 

The I oi any lamina referred to its own gravity- ^ 
axis parallel to Fis (§96) = (its mass) X Ja? 2 , and j_ 
its I v (eq. (3), §88) is .*. = its mass X ffi + fig. 115. 
its mass X 2 2 . 

Hence for the whole cone, 

I v = /* (ntfdzy - g)\^ + s 9 ] 

v 

Secondly, about a gravity-axis parallel to the base. — From 
■eq. (3), §88, with d — %h (see Prob. 7, §2*5), and the result 
just obtained, we have /= M^\f -\-\hf\. 

Thirdly, about its geometric axis, Z. — Fig. 116. Since the 
axis is perpendicular to each circular lamina through the centre, 
its 7^ (§ 97) is 

= its mass X i(rad.) 2 = (ynx*dz — g)ix 2 . 
Now x = (r -i- h)z, and hence for the whole cone 

I z = l( r xr> h- gh l )J* z'dz = <bm>hy H- g)ftfi = M&r>. 




Fig. 116. 





Fig. 118. 



102. Homogeneous Right Pyramid of Rectangular Base. — 

About its geometrical axis. Proceeding as in the last para- 



102 MECHANICS OF ENGINEERING. 

graph, we derive I z = M^d\ in which d is the diagonal of the 
base. 

103. Homogeneous Sphere. — About any diameter. Fig. 118. 
I z = ? Divide into laminae perpendicular to Z. By § 97, and 
noting that a? 2 = r 2 — z\ we have finally, for the whole sphere, 

Iz = (yx + 2g) [ \r*z - -f/V + ^) = ^ynr* -^ g 

! r 

= (i7tr a r + g)*r* = M\r>; .-. k? = %r\ 

For a segment, of one or two bases, put proper limits for z 
in the foregoing, instead of -f- r and — r. 

104. Other Cases. — Parabolic plate, Fig. 119, homogeneous 
y and of (any) constant thickness, about 
jp^v an axis through 0, the middle of the 

- J-. a ._.Jx chord, and perpendicular to the plate. 
This is 

Fig. 119. Fig. 120. I— M\(^S* + f ¥). 

The area of the segment is = %hs. 

For an elliptic plate, Fig. 120, homogeneous and of any 
constant thickness, semi-axes a and h, we have about an axis 
through O, normal to surface I = M^[a? -\- 5 2 ] ; while for a 
very small constant thickness 

I x =Mib% and I ¥ =Mia\ 

The area of the ellipse == nab. 

Considering Figs. 119 and 120 as plane figures, let the 
student determine their polar and rectangular moments of 
inertia about various axes. 

(For still other cases, see p. 518 of Kankine's Applied 
Mechanics, and pp. 593 and 594 of Coxe's Weisbach.) 

105. Numerical Substitution. — The moments of inertia of 
plane figures involve dimensions of length alone, and will be 
utilized in the problems involving flexure and torsion of beams, 
where the inch is the most convenient linear unit. E.g., the 



MOMENT OF, INERTIA. 



103 



polar moment of inertia of a circle of two inches radius about 
its centre is \nr* — 25.13 -f- biquadratic, or four-dimension, 
inches, as it may be called. Since this quantity contains four 
dimensions of length, the use of the foot instead of the inch 
would diminish its numerical value in the ratio of the fourth 
power of twelve to unity. 

The moment of inertia of a rigid body, or solid, however, 
= M ¥ = (G ■— g)Jc% in which G, the weight, is expressed in 
units of force, g involves both time and space (length), while h % 
involves length (two dimensions). Hence in any homogeneous 
formula in which the I of a solid occurs, we must be careful to 
employ units consistently ; e.g., if in substituting G -=- g for M 
(as will always be done numerically) we put g = 32.2, we 
should use the second as unit of time, and the foot as linear 
unit. 

106. Example. — Eequired the moment of inertia, about the 
axis of rotation, of a pulley consisting of a rim, four parallelo- 
pipedical arms, and a cylindrical hub which may be considered 
solid, being filled by a portion of the shaft. 
Fig. 121. Call the weight of the hub G, 
its radius T ; similarly, for the rim, G„ r x 
and r % ; the weight of one arm being = G x . 
The total / will be the sum of the Z's of 
the component parts, referred to the same 
axis, viz. : Those of the hub and rim will 
be (G ~ g)\r* and (G t - g)W? + <\ 
respectively (§ 97), while if the arms are 
not very thick compared with their length, we have for them 
(§§95 and 88) 

4 (0.-3- g) [ifa - r)* - i(r 2 - rf + [r + flr, - r)Y], 

as an approximation (obtained by reduction from the axis at 
the extremity of an arm to a parallel gravity-axis, then to the 
required axis, then multiplying by four). In most fly-wheels, 
the rim is proportionally so heavy, besides being the farthest 
removed from the axis of rotation, that the moment of inertia 
of the other parts may be for practical purposes neglected. 




Fig. 121. 



104 MECHANICS OF ENGINEERING. 

107. Ellipsoid of Inertia. — The moments of inertia about 
all axes passing through any given point of any rigid body 
whatever may be proved to be inversely proportional to the 
squares of the diameters which they intercept in an imaginary 
ellipsoid, whose centre is the given point, and whose position 
in the body depends on the distribution of its mass and the 
location of the given point. The three axes which contain the 
three principal diameters of the ellipsoid are called the Princi- 
pal Axes of the body for the given point. This is called the 
ellipsoid of inertia. (Compare § 89.) Hence the moments of 
inertia of any homogeneous regular polyedron about all gravity- 
axes are equal, since then the ellipsoid becomes a sphere. It 
can also be proved that for any rigid body, if the co-ordinate 
axes 2T, Y, and Z, are taken coincident with the three principal 
axes at any point, we shall have 

fxydM = ; fyzdM = ; and fzxdM = 0. 



DYNAMICS OF A RIGID BODY. 105 



CHAPTER Y. 

DYNAMICS OF A RIGID BODY, 

108. General Method. — Among the possible motions of a 
rigid body the most important for practical purposes (and for- 
tunately the most simple to treat) are : a motion of translation, 
in which the particles move in parallel right lines with equal 
accelerations and velocities at any given instant; and rotation 
about a fixed axis, in which the particles describe circles in 
parallel planes with velocities and accelerations proportional 
(at any given instant) to their distances from the axis. Other 
motions will be mentioned later. To determine relations, or 
equations, between the elements of the motion, the mass and 
form of the body, and the forces acting (which do not neces- 
sarily form an unbalanced system), the most direct method to 
be employed is that of two equivalent systems of forces (§ 15), 
one consisting of the actual forces acting on the body, con- 
sidered free, the other imaginary, consisting of the infinite 
number of forces which, applied to the separate material points 
composing the body, would account for their individual mo- 
tions, as if they were an assemblage of particles without mutual 
actions or coherence. If the body were at rest, then considered 
free, and the forces referred to three co-ordinate axes, they 
would constitute a balanced system, for which the six summa- 
tions 2X, -2 Y, 2Z, 2(mom.) x , 2(mom.) r , and 2(mom.) z , 
would each = ; but in most cases of motion some or all of 
these sums are equal (at any given instant), not to zero, but to 
the corresponding summation of the imaginary equivalent 
system, i.e., to expressions involving the masses of the particles 
(or material points), their distribution in. the body, and the 



106 



MECHANICS OF ENGINEERING. 



elements of the motion. That is, we obtain six equations by 
putting the 2X of the actual system equal to the 2X of the 
imaginary, and so on ; for a definite instant of time (since some 
of the quantities may be variable). 



109. Translation. — Fig. 122. At a given instant all the par- 
ticles have the same velocity = v, in parallel right lines (par- 
allel to the axis X, say), and the 
same acceleration jp. Required 
the 2X of the acting forces, 
, M shown at (I.). (II.) shows the 
imaginary equivalent system, con- 
sisting of a force = mass X ace. 
= dMp applied parallel to X to 
each particle, since such a force 
would be necessary (from eq. (IV.) 
§ 55) to account for the accelerated rectilinear motion of the 
particle, independently of the others. Putting (2Z ) x = (2X) n , 
we have 




Fig. 122. 



(2X)j =fpdM =pfdM = Mp. 



(V.) 



It is evident that the resultant of system (II.) must be paral- 
lel to X; hence that of (I.), which — (2X) X and may be de- 
noted by R, must also be parallel to X; let a = perpendicular 
distance from R to the plane YX\ a will be parallel to Z. 
Now put P(raom.) 7 ]j = [^(mom. r )]jj, (Yis an axis perpen- 
dicular to paper through 0) and we have — Ra = —fdMpz 
= —pfdMz = — pMz (§88), i.e., a = s. A similar result 
may be proved as regards y. Hence, if a rigid body has a 
motion of translation, the resultant force rn,ust act in a line 
through the centre of gravity (here more properly called the 
centre of mass), and parallel to the direction of motion. Or, 
practically, in dealing with a rigid body having a motion of 
translation, we may consider it concentrated at its centre of 
mass. If the velocity of translation is uniform, R = M X 
= 0, i.e., the forces are balanced. 




DYNAMICS OF A RIGID BODY. 107 

110. Rotation about a Fixed Axis. — First, as to the elements 
of space and time involved. Fig. 123. Let be the axis of 
rotation (perpendicular to paper), OY a fixed q 
line of reference, and OA a convenient line of 
the rotating body, passing through the axis and 
perpendicular to it, accompanying the body in 
its angular motion, which is the same as that of 
OA. Just as in linear motion we dealt with FlG - 123 - 
linear space (s), linear velocity (v), and linear acceleration (p),. 
so here we distinguish at any instant ; 
a, the angular space between Y and OA ; 

doc 

co = , , the angular velocity, or rate at which a is changing ; 

and 

„ dco d 2 a 

u = -T7 = --^r, the angular acceleration, or rate at which ft? 

is changing. 

These are all reckoned in ^-measure and may be + or — > 
according to their direction against or with the hands of a 
watch. 

(Let the student interpret the following cases : (1) at a cer- 
tain instant g? is +, and 6 — ; (2) ft? is — , and 6 +; (3) ot is 
— , &? and 6 both +; (4) a-\-, co and 6 both — .) For rotary 
motion we have therefore, in general, 



da ,^ Tv _ dco d*a 



co = 



(vi.) * = ;ir = -*?;■ ■ (vn.) 



dV K J dt ~* dt 



and .-. codco= Oda; .... . (VIII.) 

corresponding to eqs. (I.), (II.), and (III.) in § 50, for rectilinear 
motion. 

Hence, for uniform rotary motion, go being constant and 
6=0, we have a = cot, t being reckoned from the instant 
when a = 0. 

For uniformly accelerated rotary motion 6 is constant, and 



108 MECHANICS OF ENGINEERING. 

if gd denote the initial angular velocity (when a and t = 0), 

we may derive, precisely as in § 56, 

cd = oo Q + 6t ; . . (1) a = Go o t + ±df; , . (2) 

" = ^W^i • • (3) and « = i(co + GD)t. . . (4) 

If in any problem in rotary motion 6, go, and a have been 
determined for any instant, the corresponding linear values for 
any point of the body whose radial distance from the axis is p, 
will be s = ap (= distance described by the point measured 
along its circular path from its initial position), v = oop = its 
velocity, and^ = 6p its tangential acceleration, at the instant 
in question. 

Examples. — (1) "What value of &?, the angular velocity, is 
implied in the statement that a pulley is revolving at the rate 
of 100 revolutions per minute ? 

100 revolutions per minute is at the rate of 2tt X 100 
= 628.32 (^-measure units) of angular space per minute 
= 10.472 per second ; .\ ca = 628.32 per minute or 10.472 
per second. 

(2) A grindstone whose initial speed of rotation is 90 revo- 
lutions per minute is brought to rest in 30 seconds, the an- 
gular retardation (or negative angular acceleration) being con- 
stant ; required the angular acceleration, 6, and the angular 
space a described. Use the second as unit of time. 
cd 9 = 2tt|^ = 9.4248 per second ; .*. from eq. (1) 

OD — QD 

6 = —r-^ = — 9.424 -=- 30 = — 0.3141 (^-measure units) 
t 

per " square second." The angular space, from eq. (2) is 

a = Gjj + idf = 30 X 9.42 - i(0.314)900 = 141.3 

(7r-measure units), i.e., the stone has made 22.4 revolutions in 

coming to rest and a point 2 ft. from the axis has described a 

distance s = ap == 141.3 X 2 = 282.6 ft. in its circular path. 

111. Rotation. Preliminary Problem. Axis Fixed. — For 
clearness in subsequent matter we now consider the following 




DYNAMICS OF A RIGID BODY. 109 

simple case. Fig. 124 shows a rigid body, consisting of a 
drum, an axle, a projecting arm, all 
of which are imponderable, and a 
single material point, whose weight 
is G and mass M. An imponderable 
flexible cord, in which the tension is 
kept constant and = P, unwinds 
from the drum. The axle coincides 
with the vertical axis Z, while the cord ' FlG . 124> 

is always parallel to Y. Initially (i.e., when t = 0) M lies at 
rest in the plane ZY. Required its position at the end of any 
time t (i.e., at any instant) and also the reactions of the bearings 
at and O x , supposing no vertical pressure to exist at O x , and 
that P and J^are at the same level. No friction. At any in- 
stant the eight unknowns, a, go, 6, X , Y , Z , X x , and Y x , may 
be found from the six equations formed by putting 2X, etc., 
of the system of forces in Fig. 124, equal, respectively, to the 
2X, etc., of the imaginary equivalent system in Fig. 125, and 
two others to be mentioned subsequently. Since, at this in- 
stant, the velocity of M must be v = cop and its tangential ac- 
celeration p t = Op, its circular motion 
could be produced, considering it free (eq. 
(5), § 74), by a tangential force T = mass 
Xpt = MOp, and a normal centripetal 
^X--~x-^ y x force N= Mv* -*- p=M{GDpf + p= tfMp. 
fig. 125. Hence the system in Fig. 125 is equivalent 

to that of Fig. 124, and from putting the 2 (mom.) 2 of one 
= that of the other, we derive 

Pa= Tp; i.e., Pa = 6Mp\ .... (1) 

whence 6 becomes known, and is evidently constant, since P, 
a, M. and p are such. ,\ the angular motion is uniformly ac- 
celerated, and from eqs. (1) and (2), § 110, go and a become 
known ; 

i.e., Go=6t, . . . (2) and a = ^Of (3) 

Putting (2Z of 124) = {2Z of 125), gives 

Z. — G = ; i.e.. Z = G (4> 



110 MECHANICS OF ENGINEERING. 

Proceeding similarly*Kvitli the ^X of each system, 
X -f- X 1 — Tcos a — iTsin a = OMp cos a — go? Mp sin a, (5) 
and with the 2 Y of each, 
P + JT+ Y t = - Tsma-JVcosa= - dMpsina 

— <*?Mp cos a ; (6) 
while with the 2 (mom.) x we have, conceiving all the forces in 
each system projected on the plane ZJ^(see §38), and noting 
that y = p cos a and x = p sin a, 

■+ Gp cos or + Y} + P& = — {6Mp sin a)b—(co*Mp cosa)b,(7) 
:and with the -2 (mom.) r , 
— Gp sin a — X t Z = — (OMp cos ar)& -f- (oo^Mp sin a)5. . (8) 

From (7) we may find Y t ; from (8), X x ; then X and Z" 
from (5) and (6). It will be noted that as the motion proceeds 
B remains constant ; go increases with the time, cc with the 
square of the time ; Z Q is constant, = G ; while X , Y , X x , 
and 3^ have variable values dependent on p cos a and p sin <*, 
i.e., on the co-ordinates y and a? of the moving material point. 

112. Particular Supposition in the Preceding Problem with 
Numerical Substitution. — Suppose we have given (using the 
foot-pound-second system of units in which g == 32.2) G = 64.4 
lbs., whence 
M=(G + g) = 2; P = 4 lbs., I = 4ft., b = 2 ft, a = 2ft., 
and p = 4 ft.; and that M is just passing through the plane 
ZX, i.e., that a = ^7r. We obtain, first, the angular accelera- 
tion, eq. (1), 

= Pa+ Mp 2 = 8 -f- 32 = 0.25 = J. 
From eqs. (2) and (3) we have at the instant mentioned (not- 
ing that when a was == 0, I was = 0) 

co' = 2a6 == Jtt = 0.7854 +, 
while (2) gives, for the time of describing the quadrant, 
t = co -r- = 3.544. . . . seconds. 
Since at this instant cos a = and sin a = 1, we have, from 

CO, 

+ 0+r",X4 + 4X2--iX2x4X2; .-. r.= -31bs. 



DYNAMICS OF A RIGID BODY. 



Ill 



The minus sign shows it should point in a direction contrary 
to that in which it is drawn in Fig. 124. Eq. (8) gives 

- 644 x 4-X x X 4=-0+i^rx2x4x2 j.-.Xx = - 67.54 lbs. 

And similarly, knowing Y x and JT X , we have from (5) and (6), 
Xo = + 61.26 lbs., and Y = - 3.00 lbs. 
The resultant of X x and Y v also that of X , Y , and Z , can 
now be found by the parallelogram (and parallelopipedon) of 
forces, both in amount and position, noting carefully the direc- 
tions of the components. These resultants are the actions of 
the supports upon the ends of the axle ; their equals and 
ojpjposites would be the actions or pressures of the axle against 
the supports, at the instant considered (when M is passing 
through the plane ZX; i.e., with a = \7t). (At the same in- 
stant, suppose the string to break ; what would be the effect on 
the eight quantities mentioned ?) 

113. Centre of Percussion of a Rod suspended from one End. — 

Fig. 126. The rod is initially at rest (see (I.) in figure), is straight, 

homogeneous, and of constant 

(small) cross-section. Neglect its 

weight. A horizontal force or 

pressure, P, due to a blow (and 

varying in amount during the 

blow), now acts upon it from the 

left, perpendicularly to the axis, 

Z, of suspension. An accelerated 

rotary motion begins about the fixed axis Z. 




°i*- Y< 



o 



(I.) 



dT 



(in.) 



(ID 

Fig. 126. 

(II.) shows the rod 
free, at a certain instant, with the reactions X and Y Q put in 
at O . (III.) shows an imaginary system which would produce 
the same effect at this instant, and consisting of a dT = dM0p, 
and a dN = afdMfl applied to each dM, the rod being composed 
of an infinite number of dM\ each at some distance p from 
the axis. Considering that the rotation has just begun, go, the 
angular velocity is as yet small, and will be neglected. Re- 
quired Y the horizontal reaction of the support at O in terms 
of P. By putting 2 Y u = 2 Y m , we have 

P - Y =fdT= 6fpdM= 6Mp. 



112 



MECHANICS OF ENGINEERING. 



b- 



.'. To = P — OM p ; p is the distance of the centre of gravity 
from the axis (N.B. J'pdM= M p is only true when all the 
p's are parallel to each other). But the value of the angular 
acceleration 6 at this instant depends on P and a, for 2 (mom.)^ 
in (II.) = 2 (mom.)z in (III), whence Pa = 6fp*dM = 6I Z , 
where I z is the moment of inertia of the rod about Z, and from 

95 = iMl\ Now p = il; hence, finally, 

3 ar\ 
2 • I J 

Jf now Y Q is to = (X i.e., if there is to be no shock between 
the rod and axis, we need only apply P at a point whose dis- 
tance a = f I from the axis ; for then Y = 0. This point is 
called the centre of percussion for the given rod and axis. It 
and the point of suspension are interchangeable (see § 118). 
(Lay a pencil on a table; tap it at a point distant one third of 
the length from one end ; it will begin to rotate about a vertical 
axis through the farther end. Tap it at one end ; it will begin 
to rotate about a vertical axis through the point first mentioned. 
Such an axis of rotation is called an axis of instantaneous rota- 
tion, and is different for each point of impact — just as the 
point of contact of a wheel and rail is the one point of the 
wheel which is momentarily at rest, and about which, therefore, 
all the others are turning for the instant. Tap the pencil at 
its centre of gravity, and a motion of translation begins ; see 
§109.) 



114. Eotation. Axis Fixed. 



Formulae. — Consider- 





Fig. 127. 



Fig. 128. 



ing now a rigid body of ai^ shape whatever, let Fig. 127 indi- 
cate the system of forces acting at any given instant, Z being 



DYNAMICS OF A RIGID BODY. 113 

the fixed axis of rotation, oo and 6 the angular velocity and 
angular acceleration, at the given instant. X and Y are two 
axes, at right angles to each other and to Z, fixed in space. At 
this instant each dM of the body has a definite x, y, and cp 
(see Fig. 128), which will change, and also a p, and z, which will 
not change, as the motion progresses, and is pursuing a circu- 
lar path with a velocity = cop and a tangential acceleration 
±= Op. Hence, if to each dM of the body (see Fig. 128) we 
imagine a tangential force dT — dMdp and a normal force 
= dM{oopy -^ p = (rfdMp to be applied (eq. (5), § 74), and 
these alone, we have a system comprising an infinite number of 
forces, all parallel to XZ, and equivalent to the actual system 
in Fig. 127. Let 2X, etc., represent the sums (six) for Fig. 
127, whatever they may be in any particular case, while for 
128 we shall write the corresponding sums in detail. Noting 
that 

fdJV cos cp — cofdMp cos cp — oofdMy = oo*My,(§ 88); 
that/^iV sin cp — cofdMp sin cp = oofdMx — gj*Mx ; 
and similarly, that fdT cos cp = OfdMp cos cp = OMy, and 
fdT sin <p = OMx; while in the moment sums (the moment 
of dT cos cp about Y, for example, being — dT cos cp . z = 
— ddMp (cos cp)z— — OdMyz, the sum of the moms. F of all the 
{dT cos 9>)'s = - OfdMyz) 

fdT q,o% cpz = Of dMyz, fdJV sin <pz = cofdMxz, etc., 
we have, since the systems are equivalent, 

2X = + Oily - go'Mx; .... (IX.) 
2Y = - OMx- co'My ; . . . . (X.) 

2Z = 0; (XL) 

2 momB.j = - OfdMxz - cofdMyz ; . (XII.) 

2 moms. y = - OfdMyz + cofdMxz ; . (XIII.) 

2 moms. z =z OfdMp" = 0I Z . . • . (XIV.) 

These hold good for any instant. As the motion proceeds x 

and y change, as also the sums fdMxz and fdMyz. If the 

body, however, is homogeneous, and symmetrical about the 

jplane XY, fdMxz and fdMyz would always = zero; since 

9 



114 



MECHANICS OF ENGINEERING. 



the z of any dM does not change, and for every term dMy(-\-z), 
there would be a term dMy(— z) to cancel it ; similarly for 
fdMxz. The eq. (XIV.), ^ (moms, about axis of rotat.) = 
fdTp = dfdMp* = (angular accel.) X (mom. of inertia of 
body about axis of rotat.), shows how the sum fdMp 1 arises in 
problems of this chapter. That a force dT = dMBp should 
be necessary to account for the acceleration (tangential) Op of 
the mass dM, is due to the so-called inertia of the mass (§ 54), 
and its moment dTp, or OdMp*, might, with some reason, be 
called the moment of inertia of the dM, and/ '6dMp i = dfdMp 1 
that of the whole body. But custom has restricted the name 
to the sum fdM p 2 , which, being without the 6, has no term to 
suffsrest the idea of inertia. For want of a better the name is 
still retained, however, and is generally denoted by I. (See 
§§86, etc.) 



115. Example of the Preceding. — A homogeneous right par- 
Y^dfc^x allelopiped is mounted on a vertical 

axle (no friction), as in figure. is 
at its centre of gravity, hence both 
j^_ x and y are zero. Let its heaviness 
be y, its dimensions A, b lt and b (see 
§ 97). XY is a plane of symmetry, 
hence both fdMxz and fdMyz are 
zero at all times (see above). The 
tension P in the (inextensible) cord 
is caused by the hanging weight P x 
(but is not = jPj, unless the rotation is uniform). The figure 
shows both rigid bodies// 1 ^. P x will have a motion of trans- 
lation ; the parallelopiped, one of rotation about a fixed axis. 
~No masses are considered except P x -f- g. and bhb x y -s- g. The 
I z = MJc z * of the latter = its mass X T \(b' + b 2 ), § 97. At 
any instant, the cord being taut, if p = linear acceleration of 




Fig. 129. 



jP„ we have 










p = 


-Oa. ..... 


. eq. (a) 


From (XIV.), 


Pa=6I z ; 


.-. P = 6I z + a. . 


. • (1) 



DYNAMICS OF A RIGID BODY. 115 

Tor the free mass P x ~- g we have (§109) P x — P = 
mass X ace., 

= (P, + g)p = (P, 4- g)6a ; .: P = P,(l - 0a + ?). (2) 
Equate these two values of P and solve for 6, whence 

R - P ' a (*\ 

All the terms here are constant, hence 6 is constant ; there- 
fore the rotary motion is uniformly accelerated, as also the 
translation of P x . The formulae of § 56, and (1), (2), (3), and 
(4) of § 110, are applicable. The tension P is also constant ; 
see eq. (1). As for the five unknown reactions (components) 
at O x and 0„ the oearings, we shall find that they too are con- 
stant ; for 

from (IX.) we have X x + X 2 = ; (4) 

from (X.) we have P + Y x + Y % = ; (5) 

from (XI.) Ave have Z 2 - G = ; (6) 

from (XII.) we have P . AO+ Y x . ~Oft- Y % . Oft = ; (7) 
from (XIII.) we have - X x . Oft + X 3 . 6^ = 0. (8) 

Numerical substitution in the above problem,. — Let the par- 
allelopiped be of wrought-iron ; let P x — 48 lbs.; a = 6 in. = 
i ft.; 5 = 3 in. = £ ft. (see Fig. 112 ) ; 5^2 ft. 3 in. = £ ft.; 
and A = 4 in. = i ft. Also iet Oft = <9 2 6> = 18 in. = f ft., 
and ^4(9 =? 3 in. = J ft. Selecting the foot-pound-second 
system of units, in which g = 32.2, the linear dimensions must 
be used in feet, the heaviness, y, of the iron must be used in 
lbs. per cubic foot, i.e., y = 480 (see § 7), and all forces in lbs., 
times in seconds. 

The weight of the iron will be G = Vy = bbjiy = J . j- . -J- 
X 480 = 90 lbs.; its mass = 90 -f- 32.2 = 2.79 ; and its mo- 
ment of inertia about Z = I z — Mk 2 2 = M T \(b; + b*) = 2.79 
X 0.426 = 1.191. (That is, the radius of gyration, k z < = 
V0.426 = 0.653 ft.; or the moment of inertia, or any result 
depending solely upon it, is just the same as if the mass were 
concentrated in a thin shell, or a line, or a point, at a distance 
of 0.653 feet from the axis.) We can now compute the an- 
gular acceleration, 6, from eq. (3) ; 



116 MECHANICS OF ENGINEERING. 

n . 4S X j U | ■ 

~ 1.191 + (48 -f- 32.2) X i ~ 1.191 + 0.372 ~ xoo ° 
7r-measure units per " square second." The linear acceleration 
of P x is p — 6a = 7.68 feet per square second for the uniform, 
ly accelerated translation. 

Nothing has yet been said of the velocities and initial condi- 
tions of the motions ; for what we have derived so far applies 
to any point of time. Suppose, then, that the angular velocity 
gd = zero when the time, t = ; and correspondingly the ve- 
locity, v = ooa, of translation of i? 13 be also = when t = 0. 
At the end of any time t, go = 6t (§§ 56 and 110) and v =pl 
= Oat ; also the angular space, a =. %6f, described by the par- 
allel opiped during the time t, and the linear space s = ipf 
= %6af, through which the weight P x has sunk vertically. 
For example, during the first second the parallelopiped has ro- 
tated through an angle a = \6t -|X 15.36x1 = 7.68 units, 
7r-measure, i.e., (7.68 ~- 2^r) = 1.22 revolutions, while P x has 
sunk through s = \6atf = 3.84 ft., vertically. 

The tension in the cord, from (2), is 
p = 48(1 - 15.36 X i -*- g) = 48(1 - 0.24) = 36.48 lbs. 

The pressures at the bearings will be as follows, at any in- 
stant : from (4) and (8), X x and X 2 must individually be zero; 
from (6) Z 2 = Q = Vy = 90 lbs.; while from (5) and (7), Y x 
= — 21.28 lbs., and Y* = — 15.20 lbs., and should point in a 
direction opposite to that in which they were assumed in Fig. 
129 (see last lines of § 39). 

116. Torsion Balance. A Variably Accel. Rotary Motion. 

Axis Fixed. — A homogeneous solid having an axis of symmetry 

is suspended by an elastic prism, 
or filament (whose mass may be 
neglected), so that the latter i& 
vertical and coincident with the 
axis of symmetry, and is not only 
supported, but prevented from 
turning at its upper extremity. 
fig. 130. If the solid is turned about its 

axis away from its position of rest and set free, the torsional 




DYNAMICS OF A RIGID BODY. 117 

-elasticity of the rod or filament, which is fixed in the solid, 
causes an oscillatory rotary motion. Required the duration of 
an oscillation. Fig. 130. 

Take the axis Y at the middle of the oscillation (the original 
position of rest). Reckon the time from the instant of passing 
this position. Let the initial angular velocity = go . As the 
motion progresses go diminishes, i.e., 6 is negative. 

To consider the body free, conceive the rod cut close to the 
body (in which it is firmly inserted), and in the section thus 
exposed put in the vertical tension P' , and also the horizontal 
forces forming a couple to which at any instant the twisting 
action (of the portion of rod removed upon the part left in the 
free body) is known to be due. Call the moment of this couple 
Qb (known as the moment of torsion) ; it is variable, being 
directly proportional to the angle a ; hence, if by experiment 
it is found to be = Q 1 b 1 when a is = a x , for any value of a it 
will be Qb = (QJ> X -f- a t )a = Ca, in which G is the constant 
factor. 

At any instant, therefore, the forces acting are G, P ', and 
those equivalent to the couple whose moment = Qb = Col. 
■(No lateral support is required ; the student, would find the -2^, 
I 7 ,, X„ and Y 2 of Fig. 129 to be individually zero, if put in ; 
remembering that here, x and y both = 0, as also fdMxz and 
fdMyz ; and that the forces of the couple will not be repre- 
sented in any of the six summations of § 114, except in 
^ moms. z ) 

From eq. (XIV.), § 114, we have — Qb, i.e., — Got, = 6I Zy 
from which 

6 = -(C + I z )a, or, for short, 6 = — JBa. . . (1) 

Since B is constant, and there is an initial (angular) velocity 
= go , and since the variables 6, gj, and a, in angular motion 
correspond precisely to those {p, v, and s) of rectilinear motion, 
it is evident that the present is a case of harmonic ?notion, 
already discussed in Problem 2 of § 59. Applying the results 
there obtained, since B of eq. (1) corresponds to the a of that 
problem, we find that the oscillations are isochronal, i.e., their 



118 MECHANICS OF ENGINEERING. 

durations are the same whatever the amplitude (provided the 
elasticity of the rod is not impaired), and that the duration of 
one oscillation (from one extreme position to the other) is 
t' — n -~ VlT, or finally, 




t'=n VaJ z +- Qfo (2) 

117. The Compound Pendulum is any rigid body allowed to 
oscillate without friction under the action of gravity when 
mounted on a horizontal axis. Fig. 131 shows the 
body free, in any position during the progress of 
the oscillation. C is the centre of gravity; let 00 
== s. From (XIV.), § 114, we have 2 (mom. about 
fixed axis) 

= angul. ace. X mom. of inertia. 

.'. — Gs sin a = 6I , 

and 6 = — Gs sin a -+- 1 = — Mgs sin a -j- Mh\, 

i.e., 6 = — gs sin a -f- k 2 (1) 

Hence is variable, proportional to sin a:. Let us see what 
the length I = OK, of a simple circular pendulum, must be, to 
have at this instant (i.e., for this value of a) the same angular 
acceleration as the rigid body. The linear (tangential) accelera- 
tions of K, the extremity of the required simple pendulum 
would be (§ 77) p t —<J sm a > an ^ hence its angular acceleration 
would = g sin a-^l. Writing this equal to 6 in eq. (1), we 
obtain 

l = K + * (2) 

But this is independent of a; therefore the length of the sim- 
ple pendulum having an angular acceleration equal to that of 
the oscillating body is the same in all positions of the latter,. 
and if the two begin to oscillate simultaneously from a position 
of rest at any given angle a 1 with the vertical, they will keep- 
abreast of each other during the whole motion, and hence have 



DYNAMICS OF A RIGID BODY. 



119 



the same duration of oscilktion ; which is .*. , for small ampli- 
tudes (§ 78), 



if = n VI ~g = n Vk 2 +gs, .; . . . (3) 

K is called the centre of oscillation corresponding to the given 
centre of suspension 0, and is identical with the centre of per- 
cussion (§113). 

Example. — Required the time of oscillation of a cast-iron 
cylinder, whose diameter is 2 in. and length 10 in., if the axis 
of suspension is taken 4 in. above its centre. If we use 32.2 
for g, all linear dimensions should be in feet and times in 
seconds. From § 100, we have 



M{ 



TIT +tV 



•«*) 



^TTT 



From eq. (3), § 88, 

I = I c + Ms' = JfEib- W + tl = ^X 0.170; 



03 



.-. £ o 2 = 0.170 sq. ft.; .-. t'= it V0.170 -j- (32.2 Xi) = 0.395 sec. 

118. The Centres of Oscillation and Suspension are Inter- 
changeable. — (Strictly speaking, these centres are points in the 
line through the centre of gravity perpendicular to the axis of 
suspension.) Refer the centre of oscillation K to the centre 
of gravity, thus (Fig. 132, at (I.) ) : 

Mk* MJc c * + Ms' 



Sl = l 



k c 



Ms 



Ms 



= ^- « 




Now invert the body and suspend it at K\ 

required CK^ or s 3 , to find the centre of 

oscillation corresponding to K as centre of 

suspension. By analogy from (1) we have 

s 2 = ItQ -+■ s 1 ; but from (1). Jc c * -t-s 1 = s .'. 

5 2 = s ; in other words, JT, is identical with (I.) <n.) 

O. Hence the proposition is proved. FlG - m 

Advantage may be taken of this to determine the length L 
of the theoretical simple pendulum vibrating seconds, and thus 
finally the acceleration of gravity from formula (3), § 117, viz., 



120 MECHANICS OF ENGINEERING. 

when f = 1.0 and I (now = L) has been determined experi- 
mentally, we have 

g (in ft. per sq. second) = L (in ft.) X ?r a . . . (2) 

This most accurate method of determining g at any locality 
requires the use of a bar of metal, furnished with a sliding 
weight for shifting the centre of gravity, and with two project- 
ing blocks provided with knife-edges. These blocks can also 
be shifted and clamped. By suspending the bar by one knife- 
edge on a proper support, the duration of an oscillation is com- 
puted by counting the total number in as long a period of 
time as possible; it is then reversed and suspended on the 
other with like observations. By shifting the blocks between 
successive experiments, the duration of the oscillation in one 
position is made the same as in the other, i.e., the distance be- 
tween the knife-edges is the length, I, of the simple pendulum 
vibrating in the computed time (if the knife-edges are not equi- 
distant from the centre of gravity), and is carefully measured. 
The I and if of eq. (3), § 117, being thus known, g may be com- 
puted. Professor Bartlett gives as the length of the simple 
pendulum vibrating seconds at any latitude J3 

Z(in feet) = 3.26058 - 0.008318 cos 2/?. 

119. Isochronal Axes of Suspension. — In any compound 
pendulum, for any axis of suspension, there are always three 
others, parallel to it in the same gravity plane, for which the 
oscillations are made in the same time as for the first. For 
any assigned time of oscillation t' , eq. (3), § 117, compute the 
corresponding distance CO = s of O from O; 

Mh: nXMkJ + Mf 



i.e., from t' 



Mgs Mgs 



we have s = (gt'*+27t*)± ^(gT-+±n*) - k c \ . . (1) 

Hence for a given t' , there are two positions for the axis O 
parallel to any axis through C, in any gravity-plane, on both 
sides; i.e., four parallel axes of suspension, in any gravity- 
plane, giving equal times of vibration ; for two of these axes 



DYNAMICS OF A KIGID BODY. 121 

we must reverse the body. E.g., if a slender, homogeneous, 
prismatic rod be marked off into thirds, the (small) vibrations 
will be of the same duration, if the centre of suspension is 
taken at either extremity, or at either point of division. • 

Example. — Required the positions of the axes of suspension, 
parallel to the base, of a right cone of brass, whose altitude is 
six inches, radius of base, 1.20 inches, and weight per cubic inch 
is 0.304 lbs., so that the time of oscillation may be a half- 
second. (N.B. For variety, use the inch-pound-second system 
of units, first consulting § 51.) 

120. The Fly-Wheel in Fig. 133 at any instant experiences 
a pressure P' against its crank-pin from the connecting-rod 
and a resisting pressure P" from the teeth of a spur-wheel with 



Ira' in' 



Fig. 133. 



which it gears. Its weight G acts through O (nearly), and 
there are pressures at the bearings, but these latter and G have 
no moments about the axis C (perpendicular to paper). The 
figure shows it free, P" being assumed constant (in practice 
this depends on the resistances met by the machines which D 
drives, and the fluctuation of velocity of their moving parts). 
P\ and therefore T its tangential component, are variable, 
depending on the effective steam-pressure on the piston at any 
instant, on the obliquity of the connecting-rod, and in high- 
speed engines on the masses and motions of the piston and con- 
necting-rod. Let r = radius of crank-pin circle, and a the 
perpendicular from C on P" . From eq. (XI Y.), § 114, we 
have 

Tr - P"a = ei Cr .-. 6 = (Tr - P"d) - I Cj . (1 



122 



MECHANICS OF ENGINEERING. 



as the angular acceleration at any instant ; substituting which in, 
the general equation (YIII.), § 110, we obtain 

I c Godoo = Trda — P"ada. .... (2) 

From (1) it is evident that if at any position of the crank-pin 
the variable Tr is equal to the constant P"a, is zero, and 
consequently the angular velocity oo is either a maximum or a 
minimum. Suppose this is known to be the case both at m 
and n\ i.e., suppose T, which was zero at the dead-point A, 
has been gradually increasing, till at n, Tr = P"a ; and there- 
after increases still further, then begins to diminish, until at m 
Tr again = P"a, and continues to diminish toward the dead- 
point B. The angular velocity go, whatever it may have been 
on passing the dead-point A, diminishes, since 6 is negative, 
from A to n, where it is co n , a minimum ; increases from n to 
m, where it reaches a maximum value, oo m . n and m being* 
known points, and supposing co n known, let us inquire what 
Go m will be. From eq. (2) we have 



I c gjJgj^ / Trda-P" 



ada. 



(3) 



But rda = ds r = an element of the path of the crank-pin, and 
also the " virtual velocity" of the force T, and ada = ds", an 
element of the path of a point in the pitch-circle of the fly- 
wheel, the small space through which P" is overcome in dL 
Hence (3) becomes 

Ici(™m - <»»*) =J n Tds - P" X linear arc EF. (4) 

To determine / Tds we might, by a knowledge of the vary- 
ing steam-pressure, the varying obliquity of the connecting-rod, 
etc., determine T for a number of points equally spaced along 
the curve », and obtain an approximate value of this sum by 
Simpson's Rule ; but a simpler method is possible by noting 
(see eq. (1), § 65) that each term Tds of this sum = the corre- 

^ponding term Pdx in the series / 1 Pdx, in which P = the 



DYNAMICS OF A EIGID BODY. 12£ 

effective steam-pressure on the piston in the cylinder at any in- 
stant, dx the small distance described by the piston while the 
crank-pin describes any ds, and n' and m' the positions of the 
piston (or of cross-head, as in Fig. 133) when the crank-pin is 
at n and m respectively. (4) may now be written 

Pdx - P" X linear arc EF, (5) 

from which oo m may be found as proposed. More generally, it 
is available, alone (or with other equations), to determine any 
one (or more, according to the number of equations) unknown 
quantity. This problem, in rotary motion, is analogous to that 
in §59 (Prob. 4) for rectilinear motion. Friction and the in- 
ertia of piston and connecting-rod have been neglected. As 
to the time of describing the arc wm, from equations similar to 
(5), we may determine values of go for points along nm, divid- 
ing it into an even number of equal parts, calling them <& 19 go s9 
etc., and then employ Simpson's Rule for an approximate value 

|~ m n m da 
of the sum t= J n ~ ( from e q- (VI.), §110); e.g., with. 

four parts, we would have 

r-m -^ P 1 4 2 4 1 ~1 

L< = ^(angle«^^m e as.)L-+-+-+- + ^J.(6) 

121. Numerical Example. Fly-Wheel.— (See Fig. 133 and 
the equations of § 120.) Suppose the engine is non-condensing 
and non-expansive (i.e., that P is constant), and that 

P — 5500 lbs., r = 6m. = iit, a = 2it, 

and also that the wheel is to make 120 revolutions per minute, 
i.e., that its mean angular velocity is to be 

go' = i^_o x 2tc, i.e., go' = ±7t. 

First, required the amount of the resistance P" (constant) 
that there shall be no permanent change of speed, i.e., that the 
angular velocity shall have the same value at the end of a com- 
plete revolution as at the beginning. Since an equation of the 
form of eq. (5) holds good for any range of the motion, let 



124 MECHANICS OF ENGINEERING. 

that range be a complete revolution, and we shall have zero as 
the left-hand member ; /Pdx = P X 2 f t. = 5500 lbs. X 2 ft., 
or 11,000 foot-pounds (as it may be called); while P" is un- 
known, and instead of lin. arc FPwe have a whole circumfer- 
ence of 2 ft. radius, i.e., 47r ft.; 

.-. == 11,000 -P"X4X 3.1416; whence P" = 875 lbs. 
Secondly, required the proper mass to be given to the fly- 
wheel of 2 ft. radius that in the forward stroke (i.e., while the 
crank-pin is describing its upjper semicircle) the max. angular 
velocity co m shall exceed the minimum oo n by only -^ g/, assum- 
ing (which is nearly true) that i(oo m -\- co n ) =z go'. There be- 
ing now three unknowns, we require three equations, which 
are, including eq. (5) of § 120, viz.: 
Mk c *i{GD m + Go n ){oo m - co n ) 






Jn' 



Pdx - P" X linear arc EF\ (5) 

i(™m+ <»n)= <x>'=±7r; (7) and <*> m - co n = -^co f = f it. (8) 

The points n and m are found most easily and with sufficient 
accuracy by a graphic process. Laying off the dimensions to 
scale, by trial such positions of the crank-pin are found that 
T, the tangential component of the thrust P' produced in the 
connecting-rod by the steam-pressure P (which may be resolved 
into two components, along the connecting-rod and a normal 
to itself) is =(# -=- r)P'\ i.e., is = 3500 lbs. These points will 
be n and m (and two others on the lower semicircle). The 
positions of the piston n' and m', corresponding to n and m of 
the crank-pin, are also found graphically in an obvious manner. 
"We thus determine the angle nCm to be 100°, so that linear 
arc EF= iffix X 2 ft. = ^-7t, while 

/ Pdx = 5500 lbs. X / dx = 5500XraW=5500x 0.77 ft., 

■«/n/ «/»' 

.n'rnf being scaled from the draft. 

Now substitute from (7) and (8) in (5), and we have, with 
Jc c = 2 ft. (which assumes that the mass of the fly-wheel is con- 
centrated in the rim), 



DYNAMICS OF A RIGID BODY. 



125: 



{G -4- g) X 4 X 4:7t x \r = 5500 X 0.77 - 875 X *$■*, 
which being solved for G (with ^ = 32.2 ; since we have used 
the foot and second), gives G = 600.7 lbs. 

The points of max. and min. angular velocity on the back- 
stroke may be found similarly, and their values for the fly- 
wheel as now determined ; they will differ but slightly from 
the co m and oo n of the forward stroke. Professor Cotterill says 
that the rim of a fly-wheel should never have a max. velocity 
> 80 ft. per sec; and that if made in segments, not more than 
40 to 50 feet per second. In the present example we have for 
the forward stroke, from eqs. (7) and (8), oo m -= 13.2 (^-measure 
units) per second; i.e., the corresponding velocity of the wheel- 
rim is v m = oo m a ='26.4 feet per second. 

122. Angular Velocity Constant. Fixed Axis. — If go is con- 
stant, the angular acceleration, 0, must be = zero at all times, 
which requires 2 (mom.) about the axis of 
rotation to be = (eq. (XIV.), § 114). An 
instance of this occurs when the only forces 
acting are the reactions at the bearings on 
the axis, and the body's weight, parallel to 
or intersecting the axis ; the values of these 
reactions are now to be determined for dif- 
ferent forms of bodies, in various positions fig. 134. 
relatively to the axis. (The opposites and equals of these reac- 
tions, i.e., the forces with which the axis acts upon the bearings, 
are sometimes stated to be due to the " centrifugal forces" or 
"centrifugal action," of the revolving body.) 

Take the axis of rotation for Z, then, with 6 = 0, the equa- 
tions of § 114 reduce to 




2X = - w'Mx; . 

2Y = - co'My; . 

2Z = ; . . . 
2 moms. x = — oofdMyz ; 
2 moms. r = -f- cofdMxz ; 
2 moms.^ = 0. . . . 



(IXa.) 

(Xa.) 

(Xla.y 

(Xlla.) 

(XIIL&.) 

(XlVa.) 



126 MECHANICS OF ENGINEERING. 

For greater convenience, let ns suppose the axes JTand F 
(since their position is arbitrary so long as they are perpen- 
dicular to each other and to Z) to revolve with the body in its 
uniform rotation. 

122a. If a homogeneous body have a plane of symmetry 

and rotate uniformly about any axis Z perpendicular to that 

plane {intersecting it at 0), then the acting forces are equiva- 

j__ lent to a single force, = oo^Mp, applied 

""""""""] cvt and acting in a gravity-line, but 

P £^ — -| p>^> directed away from the centre of 

r <r^^%Xp&^S<X^ gravity. It is evident that such a 
./ ~~~\^^^ force P = co^Mp, applied as stated 

FlG - 135 (see Fig. 135), will satisfy all six con- 

ditions expressed in the foregoing equations, taking X through 
the centre of gravity, so that x = ~p. For, from (IXa.), P must 
= ofMp, while in each of the other summations the left- 
hand member will be zero, since P lies in the axis of X\ and 
as their right-hand members will also be zero for the present 
body (y = ; and each of the sums fdMyz &ndfdMx2 is zero, 
since for each term dMy( -f- z) there is another dMy( — z) 
to cancel it ; and similarly, fovfdMxz), they also are satisfied ; 
Q. E. D. Hence a single point of support at will suffice to 
maintain the uniform motion of the body, and the pressure 
against it will be equal and opposite to P. 

First Example. — Fig. 136. Supposing (for greater safety) 
that the uniform rotation of 210 revolutions 
per minute of each segment of a fly-wheel is / 

maintained solely by the tension in the cor- JL <^ — 
responding arm, P ; required the value of P P~" 

if the segment and arm together weigh -^ of 
a ton, and the distance of their centre of FlG - m 

gravity from the axis is p — 20 in., i.e., = -| ft. With the foot- 
ton-second system of units, with g = 32.2, we have 

P = co'Mp = [W X 2;r] a X [^r + 32 - 2 l X # = 0.83 tons, 
or 1660 lbs. 



DYNAMICS OF A KIGID BODY. 



127 




Second Example. — Fig. 137. Suppose the uniform rotation 
of the same fly-wheel depends solely on the tension in the rim, 
required its amount. The figure shows the half- <- 
rim free, with the two equal tensions, JP', put in at 
the surfaces exposed. Here it is assumed that the 
-arms exert no tension on the rim. From § 122a we 
have 2P' = oo^Mp, where JLf is the mass of the half- p' 
rim, and p its gravity co-ordinate, which may be ob- fig. m. 
tained approximately by § 26, Problem 1, considering the rim 
as a circular wire, viz., p = 2r -f- n. 

Let M = (ISO lbs.) -=- g, with r — 2 ft. We have then 

P' = i(22) 2 (180 -ir 32.2)(4 -=- it) = 1718.0 lbs. 

(In reality neither the arms nor the rim sustain the tensions 
just computed ; in treating the arms we have supposed no duty 
done by the rim, and vice versa. The actual stresses are less, 
and depend on the yielding of the parts. Then, too, we have 
supposed the wheel to take no part in the transmission of mo- 
tion by belting or gearing, which would cause a bending of the 
arms, and have neglected its weight.) 

122b. If a homogeneous body have a line of symmetry and 
rotate uniformly about an axis parallel to it (0 being the foot 
of the perpendicular from the centre of gravity on the axis), 
then the acting forces are equivalent to a single force P 
= od'Mp, applied at and acting in a gravity-line away 

from the centre of gravity. 
Taking the axis X through the 
du centre of gravity, Z being the 
axis of rotation, Fig. 138, while 
Z' is the line of symmetry, pass 
an auxiliary plane Z'Y* parallel 
to ZY. Then the sum fdMxz 
may be written fdM(p + x')z 
which = ~pfdMz + fdMx'z. 
But fdMz = 3fz = 0, since z 
= 0, and every term dM{-\- x')z is cancelled by a numerically 




*- 



Fig. 138. 



128 



MECHANICS OF ENGINEERING. 



equal term dM{— x')z of opposite sign. Hence fdMxz = 0\. 
AhofdMyz = 0, since each positive product is annulled by an 
equal negative one (from symmetry about Z'). Since, also, 
y = 0, all six conditions in § 122 are satisfied. Q. E. D. 

If the homogeneous body is any solid of revolution whose 
geometrical axis is parallel to the axis of rotation, the forego- 
ing is directly applicable. 

122c. If a homogeneous body revolve uniformly about any 
axis lying in a plane of symmetry ', the acting forces are equiv- 
alent to a single force P = od 2 Mp, acting parallel to the grav- 
ity-line which is perpendicidar to the axis (Z), and away 
from the centre of gravity, its distance from any origin in 
the axis Z being — [ fdMxz] -f- Mp {the plane ZX being a 
gravity plane). — Fig. 139. From the position of the body we 
have p = x, and y = ; hence if a 
value (rfMp be given to P and it be- 
made to act through Z and parallel to- 
X, and away from the centre of gravity^ 
all the conditions of § 122 are satisfied 
except (Xlla.) and (XHIa.). But 
symmetry about the plane XZ makes 
fdMyz = 0. and satisfies (XII&.), and 
by placing P at a distance a —fdMxz -=- Mp from along Z 
we satisfy (XIIL*.). Q. E. D. 

Example. — A slender, homogeneous, prismatic rod, of length 
= I, is to have a uniform motion, about a ver- q, 
tical axis passing through one extremity, 
maintained by a cord-connection with a fixed p 
point in this axis. Fig. 140. Given go, cp, I, j 
(p = \l cos cp), and F the cross-section of the \ { 
rod, let s = the distance from to any dM \ 
of the rod, dM being = Fyds -r- g. The x 
of any dM = s cos cp ; its z = s . sin cp; 

.'.fdMxz = (Fy -4- g) sin cp cos cp J s^ds 

-— i(Fyl -r- g)F sin cp cos cp = \MV sin cp cos <p- 




Fig. 139. 




FlQ. 140. 



DYNAMICS OF A RIGID BODY. 129 

Hence a, —fdMxz — Mp, is = f I sin (p, and the line of ac- 
tion of P(= oo^Mp = oo' (Fyl H- g) il cos cp) is therefore 
higher up than the middle of the rod. Find the intersection 
D of G and the horizontal drawn through Z at distance a from 
(?. Determine P' by completing the parallelogram GP' , at- 
taching the cord so as to make it coincide with P'\ for this w T ilI 
satisfy the condition of maintaining the motion, when once be- 
gun, viz., that the acting forces G, and the cord-tension P' 9 
shall be equivalent to a force P = atfMp, applied horizontally 
through Z at a distance a from 0, 

123. Free Axes. Uniform Rotation. — Referring again to § 122 
and Fig. 134, let 'us inquire under what circumstances the 
lateral forces, X^ Y t , X , Y , with which the bearings press 
the axis, to maintain the motion, are individually zero, i.e., that 
the hearings are not needed, and may therefore he removed 
(except a smooth horizontal plane to sustain the body's weight), 
leaving the motion undisturbed like that of a top "asleep." 
For this, not only must 2X and 2 Y both be zero, but also 
(since otherwise X 1 and X might form a couple, or Y 1 and Y 
similarly) 2 (moms.) z and 2 (moms.) F must each = zero. The 
necessary peculiar distribution of the body's mass about the 
axis of rotation, then, must be as follows (see the equations of 
§122): 

First, x and y each = 0, i.e., the axis must he a gravity-axis. 

Secondly, fdMyz = 0, and fdMxz = 0, the origin being any- 
where on Z, the axis of rotation. 

An axis (Z) (of a body) fulfilling these conditions is called 
a Free Axis, and since, if either one of the three Principal Axes 
for the centre of gravity (see § 107) be made an axis of rotation 
(the other two being taken for X and Y), the conditions 
x = 0, y = 0, fdMxz = 0, and fdMyz = 0, are all satisfied,. 
it follows that every rigid hody has at least three free axes-, 
which are the Principal Axes of Inertia of the centre of 
gravity at right angles to each other. 

In the case of homogeneous hodies free axes can often be 
determined by inspection : e.g., any diameter of a sphere; any 



130 



MECHANICS OF ENGINEERING. 



dM dP 

*> 




Fig. 141. 



transverse diameter of a right circular cylinder through its 

centre of gravity, as well as its geometrical axis; the geomet- 
rical axis of any solid of revolution ; etc. 

124. Rotation about an Axis which has a Motion of Translation. 
— Take only the particular case where the moving axis is a 
gravity-axis. At any instant, let the 
velocity and acceleration of the axis be v 
and p ; the angular velocity and accelera- 
tion about that axis, oo and 6. Then, since 
- r P the actual motion of a dM in any dt is 
compounded of its motion of rotation 
about the gravity-axis and the motion of 
translation in common with that axis, 
we may, in forming the imaginary equiva- 
lent system in Fig. 141, consider each dM as subjected to the 
simultaneous action of dP = dMp parallel to JT, of the tan- 
gential dT = dMOp, and of the normal dJY= dM{oopf -f- p 
= co 2 dMp. Take JTin the direction of translation, Z (perpen- 
dicular to paper through 0) is the moving gravity-axis ; Y 
perpendicular to both. At any instant we shall have, then, the 
following conditions for the acting forces (remembering that 
p sin cp — y.fdMy = My = ; etc.) : 

2X =fdP -fdT sin <p -fdtfox* cp = Mp; . (1) 

2Y=fdT cos cp -fdJT Bin cp = 0; . . (2) 

2 moms. z =fdTp -fdPy = Of dMp' = 6I Z = 6MJc z % (3) 

and three other equations not needed in the following example. 
Example. — A homogeneous solid of revolution rolls {with- 
out slipping) down a rough inclined 
plane. Investigate the motion. Con- 
sidering the hodyfree, the acting forces 
are G (known) and N and P, the un- x 
known normal and tangential compo- *V 
nents of the action of the plane on the 




^ 



Fig. 142. 



roller. If slipping occurs, then P is the 

sliding friction due to the pressure iV(§ 156); here, however, it is 



DYNAMICS OF A RIGID BODY. 131 

less by hypothesis (perfect rolling). At any instant the four 
unknowns are found by the equations 

2I,i.e, Ganfi- P, = (G+g)p\ . (1) 
2Y,i.e., Gcos/3 - if, = 0; .... (2) 
2 moms. z , i.e., Pa, = dMk z * ; . . (3) 

while on account of the perfect rolling, 

6a = p (4) 

Solving, we have, for the acceleration of translation, 

p = gsmp+[l + (]c z * + a>)l 
(If the body slid without friction, jp would = g sin /3.) Hence 
for a cylinder (§ 97), Jc z being = J& a , we have^> = \g sin ytf ; 
and for a sphere (§ 103) p = \g sin /?. 

(If the platie is so steep or so smooth that both rolling and 
slipping occur, then da no longer = p, but the ratio of P to N 
is known from experiments on sliding friction ; hence there are 
still four equations.) 

The motion of translation being thus found to be uniformly 
accelerated, we may use the equations of § 56 for finding dis- 
tance, time, etc. 

Query. — How may we distinguish two spheres by allowing 
them to roll down the same inclined plane, if one of them is 
silver and solid, while the other is of gold, but silvered and 
hollow, so as to be the same as the first in diameter, weight, 
and appearance? 

125. Parallel-Rod of a Locomotive. — When the locomotive 
moves uniformly, each dM of the rod between the two (or 
three) driving-wheels rotates with j \ ; 

uniform velocity about a centre of its 9 B I I \ P $ tr I 
own on the line BD, Fig. 143, and \^\y a) X^f^/ 
with a velocity v and radius r common ^^^ 
to all, and likewise has a horizontal ( k\ ; t A jj 
uniform motion of translation. Hence en.) 

if we inquire what are the reactions P FlG - 143 - 

of its supports, as induced solely by its weight and motion, 
when in its lowest position (independently of any thrust along 



132 MECHANICS OF ENGINEERING. 

the rod), we put 2T of (I.) = 2Y of (II.) (II. shows the 

imaginary equivalent system), and obtain 

2P - G =fdN =fdMv 2 + r = (v* + r)fdM = Mv % -r- r. 

Example. — Let the velocity of translation = 50 miles per 
hour, the radius of the pins be 18 in. = f ft., and = half that 
of the driving-wheels, while the weight of the rod is 200 lbs. 
With g = 32.2, we must use the foot and second, and obtain 

v = £[50 X 5280 -f- 3600] ft. per second = 36.6; 

while M= 200 + 32.2 = 200 X .0310 = 6.20 ; 

and finally P = i[200 + 6.2(36.6) 2 -^- f] = 2868.3 lbs., 

or nearly \\ tons, about thirty times that due to the weight 
alone. 

126. So far in this chapter the motion has been prescribed, 
and the necessary conditions determined, to be fulfilled by the 
acting forces at any instant. Problems of a converse nature, 
i.e., where the initial state of the body and the acting forces 
are given while the resulting motion is required, are of much 
greater complexity, but of rare occurrence in practice. The 
reader is referred to Rankine's Applied Mechanics. A treat- 
ment of the Gyroscope will be found in the American Journal 
of Science for 1857, and in the article of that name in Johnson's 
Cyclopaedia. 



WORK, ENERGY, AND POWER. 133 



CHAPTER VI. 

WORK, ENERGY, AND POWER. 

127. Remark. — These quantities as defined and developed 
in this chapter, though compounded of the fundamental ideas 
of matter, force, space, and time, enter into theorems of such 
wide application and practical use as to more than justify their 
consideration as. separate kinds of quantity. 

128. Work in a Uniform Translation. Definition of Work. — 

Let Fig. 144 represent a rigid body having a motion of trans- 
lation parallel to X, acted on by a 
system of forces P x , P„ P 3 , and i? 4 , 
which remain constant. 

Let 8 be any distance described by 
the body during its motion ; then ~2X 
must be zero (§ 109), i.e., noting that 
P 3 and i? 4 have negative X com- 
ponents (the supplements of their 
angles with Jf are used), 

P x cos a x -f- P a cos <* a — P 3 cos a s — P 4 cos a t — ; 

or, multiplying by s and transposing, we have (noting that 
s cos a x = s x the projection of s on P x , that s cos or 8 = s 2 , the 
projection of s on P 2 , and so on), 

PA + PA = ^.*. + ^A (a) 

The projections s x , s 2 , etc., may be called the distances de- 
scribed in their respective directions by the forces P x , P„ etc.; 
P 1 and P 9 having moved forward, since s x and s 2 fall in front 
of the initial position of their points of application ; P 3 and P t 
backward, since s 3 and s 4 fall behind the initial positions in 
their case. (By forward and backward we refer to the direc- 



kR 3 


__ — ^ //\Vi 


—}-<:;--a~~7 " v 




Fig. 144. 



134 MECHANICS OF ENGINEERING. 

tion of each force in turn.) The name Work is given to the- 
product of a force by the distance described in the direction 
of the force by the point of application. If the force moves 
forward (see above), it is called a working-force, and is said to- 
do the work (e.g., P r s^) expressed by this product; while if 
backward, it is called a resistance, and is then said to have the 
work (e.g., P 3 s 3 ), done upon it, in overcoming it through the 
distance mentioned (it might also be said to have done nega- 
tive work). 

Eq. (a) above, then, proves the theorem that : In a uniform 
translation, the working forces do an amount of work which 
is entirely applied to overcoming the resistances. 

129. Unit of Work. — Since the work of a force is a product 
of force by distance, it may logically be expressed as so many 
foot-pounds, inch-pounds, kilogram-meters, according to the 
system of units employed. The ordinary English unit is the 
foot-pound, or ft.-lb. It is of the same quality as a force- 
moment. 

130. Power. — Work as already defined does not depend on 
the time occupied, i.e., the work P 1 s l is the same whether per- 
formed in a long or short time ; but the element of time is of 
so great importance in all the applications of dynamics, as well 
as in such practical commercial matters as water-supply, con- 
sumption of fuel, fatigue of animals, etc., that the rate of work 
is a consideration both of interest and necessity. 

Power is the rate at which work is done, and one of its 
units is one foot-pound per second in English practice ; a larger 
one will be mentioned presently. 

The power exerted by a working force, or expended upon a 
resistance, may be expressed symbolically as 

L = P 1 s 1 -r- t, or i? s s 3 ^ t, 

in which t is the time occupied in doing the work P& or JR 3 s t 
(see Fig. 144) ; or if v y is the component in the direction of 
the force P, of the velocity v of the body, we may also write 

L = P\% (by 




WORK, ENERGY, AND POWER. 135 

131. Example. — Fig. 145, shows as a free body a sledge 
which is being drawn uniformly up N ^%*p^ 

a rough inclined plane by a cord 
parallel to the plane. Required the 
total power exerted (and expended), 
if the tension in the cord is P x — 100 
lbs., the weight of sledge i? 3 = 160 " Fig. 145. 

lbs., /3 = 30°, and the sledge moves 210 ft. each minute. N 
and JB 4 are the normal and parallel (i.e., i? 4 ==: friction) com- 
ponents of the reaction of the plane on the sledge. From eq. 
(1), § 128, the work done while the sledge advances through 
s = 210 ft. may be obtained either from the working forces, 
which in this case are represented by jP, alone, or from the 
resistances P 3 and 7? 4 . Take the former method first. Pro- 
jecting s upon P x we have s x = s. 
Hence P^ or 100 lbs. X 240 ft. = 24,000 ft.-lbs. 
of work done in 60 seconds. That is, the power exerted by the 
working forces is 

L = P l s 1 -T- t — 400 ft.-lbs. per second. 

As to the other method, we notice that M 3 and i? 4 are resist- 
ances, since the projections s 3 = s sin /?, and s 4 = *, would fall 
back of their points of application in the initial position, while 
JV is neutral, i.e., is neither a working force nor a resistance, 
since the projection of s upon it is zero. 

From 2X= we have — E 4 — i? 3 sin j3 + P x = 0, 

and from 2 Y = (§ 109) N - E z cos p =0; 

whence 7?, the friction =i 20 lbs., and N = 138.5 lbs. Also, 
since s, — s sin p = 240 X i = 120 ft., and s A = s, = 240 ft., 
we have for the work done upon the resistances (i.e., in over- 
coming them) in 60 seconds 

i? 3 6- 3 + R A s< =t 160 X 120 + 20 X 240 = 24,000 ft.-lbs., 

and the power expended in overcoming resistances ^ 

L = 24,000 -=- 60 = 400 ft.-lbs. per second, 
as already derived. Or, in words the power exerted by the 



136 MECHANICS OF ENGINEERING. 

tension in the ccd is expended entirely in raising the weight 
a vertical height of 2 feet, and overcoming the friction through 
a distance of 4 feet along the plane, every second ; the motion 
being a uniform translation. 

132. Horse-Power. — As an average, a horse can exerts a trac- 
tive effort or pull of 100 lbs., at a uniform pace of 4 ft. per sec- 
ond, for ten hours a day without too great fatigue. This gives 
a power of 400 f t.-lbs. per second ; but Boulton & Watt in 
rating their engines, and experimenting with the strong dray- 
horses of London, fixed upon 550 ft.-lbs. per second, or 33,000 
ft.-lbs. per minute, as a convenient large unit of power. (The 
French horse-power, or cheval-vapeur, is slightly less than the 
English, being 75 kilogram meters per second, or 32,550 ft.-lbs. 
per minute.) This value for the horse-power is in common 
use. In the example in § 131, then, the power of 400 ft.-lbs. 
per second exerted in raising the weight and overcoming fric- 
tion may be expressed as (400-=- 550 =) -^ of a horse-power. A 
man can work at a rate equal to about -^ of a horse-power, 
with proper intervals for eating and sleeping. 

133. Kinetic Energy. Retarded Translation. — In a retarded 
translation of a rigid body whose mass = M, suppose there 
are no working-forces, and that the resistances are constant and 
their resultant is R. (E.g., Fig. 146 shows such a case ; a 

^ N sledge, having an initial velocity c and slid- 

—7 >v ing on a rough horizontal plane, is gradu- 



al 
R 



MfflW/Ww///// ' ally retarded by the friction R.) R is par- 

- G allel to the direction of translation (§ 109) 

fig. 146. and the acceleration is ]) = — R -=- M ; 

hence from vdv =pds we have 

/vdv = - (1 + M)/Bds (1) 

But the projection of each ds of the motion upon R is = ds 
itself ; i.e. (§ 128), Rds is the work done upon R, in overcom- 
ing it through the small distance ds, and /Rds is the sum of 
all such amounts of work throughout any definite portion of 
the motion. Let the range of motion be between the points 



WORK, ENERGY, AND POWER. 137 

•where the velocity = c, and where it = zero (i.e., the mass 
has come to rest). With these limits in eq. (1) (0 and s' be- 
ing the corresponding limits for s), we have 

That is, in giving up all its velocity c the body has been able 
to do the work/Eds (this, if B remains constant, reduces to 

Bs') or its equal — ~— . If, then, by energy we designate the 

ability to perform work, we give the name kinetic energy of 
a moving body to the product of its mass by half the square 

(Mv 2 \ 
of its velocity i^o~~ J; i.e., energy due to motion. (The anti- 
quated term vis viva was once applied to the form Mv 2 .) 

134. Work and Kinetic Energy in any Translation. — Let P 

be the resultant of the working forces at any instant, B that 
of the resistances ; they (§ 109) will both M 

act in a gravity-line parallel to the di- < & 

rection of translation. The acceleration 0« -s.'- >d . 

at any instant is p = {^2X -~- M) FlG# 147# 

= (P — B) — M; hence from vdv — pds we have 

Mvdv = Pds — Bds (1) 

Integrating between any two points of the motion as and O r 
where the velocities are v and v\ we have after transposition 



t« 



/~ -/-+£?- ^l ■ 



(d) 



But P being the resultant of P 1? P n etc., and B that of 
R x , B„ etc., we may prove, as in § 62, that if du x , du„ etc., be 
the respective projections of any ds upon P v P„ etc., while 
dw t , dw„ etc., are those upon B^ B 2 , etc., then 

Pcls—P^du^P^du^ and Bds=B l dw 1 -\-B i dw, ; 

and (d) may be rewritten 



138 MECHANICS OF ENGINEERING. 

= fji t dw, +£'R,dw, + .... + [~ - ^f'] ; (4 

or, in words : In any translation, a portion of the work done 
by the working forces is applied in overcoming the resistances 
while the remainder equals the change in the kinetic energy of 
the body. 

It will be noted that the bracket in (e) depends only on the 
initial and final velocities, and not upon any intermediate 
values ; hence, if the initial state "is one of rest, and also the 
final, the total change in kinetic energy is zero, and the work 
of the working forces has been entirely expended in the work 
of overcoming the resistances ; but at intermediate stages the 
former exceeds the work so far needed to overcome resistances, 
and this excess is said to be stored in the moving mass ; and as 
the velocity gradually becomes zero, this stored energy becomes 
available for aiding the working forces (which of themselves 
are then insufficient) in overcoming the resistances, and is then 
said to be restored. (The function of a fly-wheel might be 
stated in similar terms, but as that involves rotary motion it 
will be deferred.) 

Work applied in increasing the kinetic energy of a body is 
sometimes called " work of inertia," as also the work done by 
a moving body in overcoming resistances, and thereby losing 
speed. 

135. Example of Steam-Hammer. — Let us apply eq. (e) to 
determine the velocity v' attained by a steam-hammer at the 
lower end of its stroke (the initial velocity being = 0), just 
before delivering its blow upon a forging, supposing that 
the steam-pressure P^ at all stages of the downward stroke is 
given by an indicator. Fig. 148. Weight of moving mass 
is 322 lbs.; .'. M = 10 (foot-pound-second system), I = 1 foot. 
The working forces at any instant are P x = G = 322 lbs.; P„ 
which is variable, but whose values at the seven equally spaced 



WORK, ENERGY, AND POWER. 



139 



XZJ 



a 

h 

c 

d 

e 

-f 

hflf 



Fig. 148. 



points a, b, c, d, e, f, g, are 800, 900, 900, 800, 600, 500, 450 

lbs., respectively. R, the exhaust-pressure (16 

lbs. per sq. inch X 20 sq. inches piston-area) = 

320 lbs., is the only resistance, and is constant. 

Hence from eq. (<?), since here the projections 

du„ etc., of any ds upon the respective forces 

are equal to each other and = ds, 

PJ ds + J P 2 ds = RJ ds + ^~. (1) 

The term fP^ds can be obtained approximately 
by Simpson's Rule, using the above values for 
six equal divisions, which gives 

-^[800'+ 4(900 + 800 + 500) 
+ 2(900 + 600) +450] 
= 725 ft.-lbs. of work. Hence, making all the substitutions^ 

we have, since / ds = 1 ft., 

322 X 1 + 725 = 320 X 1 + \Mv n \ •'• $Mv n = 727 ft.-lbs. 
of energy to be expended in the forging. (Energy is evi- 
dently expressed in the same kind of unit as work.) We may 
then say that the forging receives a blow of 727 ft.-lbs. 
energy. The pressure actually felt at the surface of the ham- 
mer varies from instant to instant during the compression of 
the forging and the gradual stopping of the hammer, and 
depends on the readiness with which the hot metal yields. 

If the mean resistance encountered is R m , and the depth of 
compression s", we would have (neglecting the force of gravity, 
and noting that now the initial velocity is v\ and the final 
zero), from eq. (c), 

%Mv' 2 = R m s" ; i.e., R m = [727 -r- s" (ft.)] lbs. 

E.g., if s" — -§- of an inch = -fa of a foot, R m = 43620 lbs., 
and the maximum value of R would probably be about double 
this near the end of the impact. If the anvil also sinks during 
the impact a distance s"\ we must substitute s f " + s" instead 
of s" ; this will give a smaller value for R m . 



140 



MECHANICS OF ENGINEERING. 



By mean value for R is meant [eq. (c)~\ that value, R m} which 
satisfies the relation 



R r 



f Eds. 



This may be called more explicitly a space-average, to dis- 
tinguish it from a time-average, which might appear in some 
problems, viz,, a value R tm , to satisfy the relation (f being the 
duration of the impact) 

jfttmt' =J Rdt, 

and is different from R m . 

From iMv" = 727 ft.-lbs., we have v' = 12.06 ft. per sec, 
whereas for a free fall it would have been V^X 32.2x1 = 8.03. 
(This example is virtually of the same kind as Prob. 4, § 59, 
differing chiefly in phraseology.) 



136. Pile-Driving. — The safe load to be placed upon a pile after 
the driving is finished is generally taken as a fraction (from -$• 
to ■§) of the resistance of the earth to the passage of the pile as 
indicated by the effect of the last few blows of the ram, in ac- 
cordance with the following approximate theory : Toward the 
t end of the driving the resistance R encountered by 

the pile is nearly constant, and is assumed to be that 
^ met by the ram at the head of the pile ; the distance 

s' through which the head of the pile sinks as an 
\hWfh effect of the last blow is observed. If G, then, is 
the weight of the ram, = Mg, and h the height of 
free fall, the velocity due to A, on striking the pile, 
is c = V2gh (§ 52), and we have, from eq. (c), 



:: --w 



Fig. 149 



\Mc\ i.e., Gh, =f S Rds = Rs f 



(1) 



(R being considered constant) ; hence R = Gh -f- s', 
and the safe load (for ordinary wooden piles), 

P = from £ to i of Gh -f- s' (2) 

Maj. Sanders recommends \ from experiments made at Fort 



WORK, ENERGY, AND POWER. 141 

Delaware in 1851; Molesworth, £; General Barnard, -J-, from 
extensive experiments made in Holland. 

Of course from eq. (2), given P, we can compute s'. 

(Owing to the uncertainty as to how much of the resistance 
R is due to friction of the soil on the sides of the pile, and 
how much to the inertia of the soil around the shoe, the more 
elaborate theories of Weisbach and Ran kin e seem of little 
practical account.) 

137. Example. — In preparing the foundation of a bridge-pier 
it is found that each pile (placing them 4 ft. apart) must bear 
safely a load of 72 tons. If the ram weighs one ton, and falls 
12 ft., what should be the effect of the last blow on each pile? 
Using the foot-ton-second system of units, and Molesworth 
factor -J-, eq. (2) gives 

s' = |(1 X 12 ^ 72) = ^L- of a foot = J of an inch. 

That is, the pile should be driven until it sinks only \ inch 
under each of the last few blows. 

138. Kinetic Energy Lost in Inelastic Direct Central Impact. — 

Referring to § 60, and using the same notation as there given, 
we find that if the united kinetic energy possessed by two in- 
elastic bodies after their impact, viz., ^M X C* -f- -JJ^C 2 , Shav- 
ing the value {M x c x + J/> 2 ) -r- {M x + M t ), be deducted from 
the amount before impact, viz., \M x c x -f- -JJf/^ 2 , the loss of 
kinetic energy during impact of two inelastic bodies is 

w ~ M x + m}° 1 g * } W 

An equal amount of energy is also lost by partially elastic 
bodies during the first period of the impact, but is partly re- 
gained in the second. If the bodies were perfectly elastic, we 
would find it wholly regained and the resultant loss zero, from 
the equations of § 60 ; but this is not quite the reality, on 
account of internal vibrations. 

The kinetic energy still remaining in two inelastic bodies 
after impact (they move together as one mass) is 



142 



MECHANICS OF ENGINEERING. 



%{M X -f- -^Q^ 2 ? or > after inserting the value of 
C = (M x c x + M t c,) -r- (M x + JQ, we have 



W = - 

2 



1 [Jf^+Jfo]' 



J* x + M t 



(2) 



Example 1. — The weight 6^ = J/"^ falls freely 
through a height A, impinging upon a weight G % 
= -3/,^, which was initially at rest. After their {in- 
elastic) impact they move on together with the com- 
bined kinetic energy just given in (2), which, since 
c x and <? 2 , the velocities before impact, are respectively 
V2gh and 0, may be reduced to a simpler form. 
This energy is soon absorbed in overcoming the 
flange-pressure R, which is proportional (so long as 
the elasticity of the rod is not impaired) to the 
elongation s, as with an ordinary spring. If from 
Fig. 150. previous experiment it is known that a force R 

produces an elongation s oy then the variable R = (R -f- s )s. 

Neglecting the weight of the two bodies as a working force, 

we now have, from eq. (d), 



ds 







R n 



R. 



= —f sds + 0- 
M?gh 



M?gh 



; * til — — 1 y f<*\ 

Le '' s * 2 - M x +M % W 

When s = s\ i.e., when the masses are (momentarily) at rest 
in the lowest position, the flange-pressure or tensile stress in the 
rod is a maximum, R' — (R ~ s )s', whence s' = R's -f- R ; 
and (3) may be written 

M*gh 



or 



2 

R f \ 



8 = 



M,'gh 



(4) 



(5) 



Eq. (3) gives the final elongation of the rod, and (5) the greatest 
tensile force upon it, provided the elasticity of the rod is not 



WORK, ENERGY, AND P0A7ER. 143 

impaired. The form £2?V in (4) may be looked upon as a direct 

integration of / Rds, viz., the mean resistance (-Ji?') multi- 

plied by the whole distance (V) gives the work done in over- 
coming the variable R through the successive ds's. 

If the elongation is considerable, the working-forces G 1 and 
G^ cannot be neglected, and would appear in the term -|- (6?! 
+ #,)*' in the right-hand members of (3 s ), (4), and (5). The 
upper end of the rod is firmly fixed, and the rod itself is of 
small mass compared with M x and M v 

Example 2. — Two cars, Fig. 151, are connected by an elastic 
chain on a horizontal track. Velocities before impact (i.e., 

before the stretching of the chain be- >c 2 o km 

gins, by means of which they are I4^___^__| 
brought to a common velocity at the M 2 Mx 

instant of greatest tension R ', and Fig. m. 

elongation s' of the chain) are c x = e v and e 2 == 0. 

During the stretching, i.e., the first period of the impact, the 
kinetic energy lost by the masses has been expended in stretch- 
ing the chain, i.e., in doing the work Ji?V ; hence we may 
write (the elasticity of the chain not being impaired) (see eq. (1) ) 

M x Mtf _ 1 _ R s»_ITs 

^M.+M," 2 ~ s ' 2~ 2R ~> ' ' W 

in which the different symbols have the same meaning as in 
Example 1, in which the rod corresponds to the chain of this 
example. 

(Let the student explain why the stipulation is not made here 
that one end of the chain shall remain fixed.) 

In numerical substitution, 32.2 for g requires the use of the 
units foot and second for space and time, while the unit of 
force may be anything convenient. 

139. Work and Energy in Rotary Motion. Axis Fixed. — 

The rigid body being considered free, let an axis through O 
perpendicular to the paper be the axis of rotation, and resolve 
all forces not intersecting the axis into components parallel 




144 MECHANICS OF ENGINEERING. 

and perpendicular to the axis, and the latter again into com- 
ponents tangent and normal to the circular path of the point 

of application. These tangential com- 
ponents are evidently the only ones 
of the three sets mentioned which 
have moments about the axis, those 
having moments of the same sign as 
od (the angular velocity at any instant) 
being called working forces, T x , T 2 , 
etc. ; those of opposite sign, resist-- 
FlG - 152 - ances, T/, T/, etc.; for when in time 

dt the point of application JB^ of T x , describes the small arc 
ds^ = a x da, whose projection on T x is = ds x , this projection 
falls ahead (i.e., in direction of force) of the position of the 
point at the beginning of dt, while the reverse is true for T/* 
From eq. (XIY.), § 114, we have for 6 (angul. accel.) 

6 = j , (1) 

which substituted in codoo = 6da (from § 110) gives (remem- 
bering that a x da = ds x , etc.), after integration and transposition, 

/n n,n 

T t ds, +J T,ds, + etc. 

= £ TM +£ T,'ds,' + etc. + H».'/ - i«.'7], (2) 

where and n refer to any two (initial and final) positions of 
the rotating body. Eq. (4), § 120, is an example of this. 

Now \oo^I — ^GDnfdMp* =fidM(Go n p)% which, since oo n p 
is the actual velocity of any dM *X, this (final) instant, is nothing 
more than the sum of the amounts of kinetic energy possessed 
at this instant by all the particles of the body ; a similar state- 
ment may be made for iooJI. 

Eq. (2) therefore may be put into words as follows: 

Between any two positions of a rigid body rotating about a 

fixed axis, the work done by the working forces is partly used 

in overcoming the resistances, and the remainder in changing 

the kinetic energy of the individual particles. If in any case 



WORK, ENERGY, AND POWER. 145 

this remainder is negative, the final kinetic energy is less than 
the initial, i.e., the work done by the working forces is less than 
that necessary to overcome the resistances through their respec- 
tive spaces, and the deficiency is made up by the restoring of 
some of the initial kinetic energy of the rotating body. A 
moving fly-wheel, then, is a reservoir of kinetic energy. 

Eq. (2) has already been illustrated numerically in § 121, 
where the additional relation was utilized (for a connecting-rod 
and piston of small mass), that the work done in the steam- 
cylinder is the same as that done directly at the crank-pin by 
the working-force there. 

140. Work of Equivalent Systems the Same. — If two plane 
systems of forces acting on a rigid body are equivalent (§ 1 5a), 
the aggregate work done by either of them during a given slight 
displacement or motion of the body parallel to their plane is 
the same. By aggregate work is meant what has already been 
defined as the sum of the " virtual moments" (§§ 61 to 64), in 
any small displacement of the body, viz., the algebraic sum of 
the products, 2 {Pdu), obtained by multiplying each force by 
the projection (du) of the displacement of (or small space 
described by) its point of application upon the force. (We 
here class resistances as negative working forces.) 

Call the systems A and B ; then, if all the forces of B were 
reversed in direction and applied to the body along with those 
of A, the compound system would be a balanced system, and 
hence we would have (§ 64), for a small motion parallel to the 
plane of the forces, 

2{Pdu) = 0, i.e., 2(PM) for A - 2(Pdu) for B = 0, 

or + 2{Pdu) for A = + 2(Pdu) for B. 

But -f- ^ (Pdu) for A is the aggregate work done by the forces 
of A during the given motion, and -f- ^(Pdu) for B is a 
similar quantity for the forces of B (not reversed) during the 
same small motion if B acted alone. Hence the theorem is 
proved, and could easily be extended to space of three dimen- 
sions. 

10 



146 



MECHANICS OF ENGINEERING. 




Fig. 153. 

0, of the body ; a final, n 



141. Relation of Work and Kinetic Energy for any Extended 
Motion of a Rigid Body Parallel to a Plane. — (If at any instant 

any of the forces acting are not 
parallel to the plane mentioned, 
their components lying in or 
parallel to that plane, will be used 
instead, since the other compo- 
nents obviously would be neither 
working forces nor resistances.) 
Fig. 153 shows an initial position, 
and any intermediate, as q. The 
forces of the system acting may vary in any manner during 
the motion. 

In this motion each dM describes a curve of its own with 
varying velocity v, tangential acceleration p t , and radius of 
curvature r ; hence in any position q, an imaginary system B 
(see Fig. 154), equivalent to the actual system A (at q in Fig. 
153), would be formed by applying to each dM a 
tangential force dT = dMp ty and a normal force , 
dJV = dMv 2 -f- r. By an infinite number of con- 
secutive small displacements, the body passes from 
o to n. In the small displacement of which q is the 
initial position, each dM describes a space ds, and 
dT does the work dTds = dMvdv, while dJV does the work- 
dN X = 0. Hence the total work done by B in the small 
displacement at q would be 



clT 



dM'v'dv' + dM"v"dv" + etc., 



(1) 



including all the dM 's of the body and their respective veloci- 
ties at this instant. 

But the work at q in Fig. 153 by the actual forces (i.e., of 
system A) during the same small displacement must (by § 140) 
be equal to that done by B, hence 

P x du, + P 3 du, + etc. = dM'v'dv' + dM"v"dv" + etc. (g) 

Now conceive an equation like (q) written out for each of 



WORK, ENERGY, AND POWER. 147 

the small consecutive displacements between positions o and 
n and corresponding terms to be added ; this will give 

J n P x du x +f n p,du, + etc. 

= dM'f n v'dv' + dM"f n <o"dv" + etc. 
= idM'(v n " - O + idM ,, (v n ,n - v fn ) + etc. 
The second member may be rewritten so as to give, finally, 

J n P x du x +f n P^+etc. = 2(idMv n 2 )-2(idMv *), (XY.) 

or, in words, the work done by the acting forces (treating a re- 
sistance as a negative working force) between any two posi- 
tions is equal to the gain (or loss) in the aggregate kinetic 
energy of the particles of the body between the two positions. 
To avoid confusion, 2 has been used instead of the sign y* in 
one member of (XY.), in which v n is the final velocity of any 
dM (not the same for all necessarily) and v the initial. 

(The same method of proof can be extended to three dimen- 
sions.) 

Since kinetic energy is always essentially positive, if an ex- 
pression for it comes out negative as the solution of a problem, 
some impossible conditions have been imposed. 

142. Work and Kinetic Energy in a Moving Machine. — 

Defining a mechanism or machine as a series of rigid bodies 
jointed or connected together, so that working-forces applied 
to one or more may be the means of overcoming resistances 
occurring anywhere in the system, and also of changing the 
amount of kinetic energy of the moving masses, let us for 
simplicity consider a machine the motions of whose parts are 
all parallel to a plane, and let all the forces acting on any one 
piece, considered free, at any instant be parallel to the same 
plane. 

Now consider each piece of the machine, or of any series of 
its pieces, as a free body, and write out eq. (XY.) for it be- 
tween any two positions (whatever initial and final positions are 



148 



MECHANICS OF ENGINEERING. 



selected for the first piece, those of the others must be corre- 
sponding initial and corresponding final positions), and it will 
be found, on adding up corresponding members of these equa- 
tions, that the terms involving those components of the mutual 
pressures (between the pieces considered) which are normal 
to the rubbing surfaces at any instant will cancel out, while 
their components tangential to the rubbing surfaces (i.e., fric- 
tion, since if the surfaces are perfectly smooth there can be 
no tangential action) will appear in the algebraic addition as 
resistances multiplied by the distances rubbed through, meas- 
ured on the rubbing surfaces. For example, Fig. 155, where- 
one rotating piece both presses and rubs on another. Let the- 
normal pressure between them at A be i? 2 = B 2 ; it is a work- 
ing force for the body of mass M" , but a resistance for M ', 
hence the separate symbols for the numerically equal forces 
(action and reaction). 

Similarly, the friction at A is B 3 = B s ; a resistance for M', 
a working-force for M" . (In some cases, of course, friction 
may be a resistance for both bodies.) For a small motion, A 
describes the small arc AA' about 0' in dealing with M' , but 
for M" it describes the arc AA" about O' 1 ', A' A" being 
parallel to the surface of contact AD, while AB is perpen- 




Fig. 156. 



Fig. 157. 



Fig. 155. 

dicular to A' A" . In Figs. 156 and 157 we see W and M" 
free, and their corresponding small rotations indicated. During 
these motions the kinetic energy (K. E.) of each mass has 
changed by amounts d(K. E.)^/ and d(K. ~E.)m" respectively, and 
hence eq. (XV.) gives, for each free body in turn, 

P~a~ti - B,AB - B.AJB = d(K. E. V • (1> 

- B'bW 7 + P.AB + P,A^B = d(K. E. V. . (2) 



WORK, ENERGY, AND POWER. 149 

[Now add (1) and (2), member to member, remembering that 
jP 2 = i? 2 and P 3 = R z — F z = friction, and we have 



P x aa' - F Z A'A" - R$b" = d(K. E.) M , + d(K. E.) M „, (3) 

in which the mutual actions of M' and 1£" do not appear, 
except the friction, the work done in overcoming which, when 
the two todies are thus considered collectively, is the product 
of the friction by the distance A' A" of actual rubbing meas- 
ured on the rubbing surface. For any number of pieces, then, 
considered free collectively, the assertion made at the beginning 
of this article is true, since any finite motion consists of an 
infinite number of small motions to each one of which an equa- 
tion like (3) is applicable. 

Summing the corresponding terms of all such equations, we 
have 

f n P x du, + f n p,du,+ etc. = 2CK.E.),- 2(K.E.),(XVI.) 

This is of the same form as (XT.), but instead of applying to a 
single rigid body, deals with any assemblage of rigid parts 
forming a machine, or any part of a machine (a similar proof 
w T ill apply to three dimensions of space); but it must be remem- 
bered that it excludes all the mutual actions of the pieces con- 
sidered except friction, which is to be introduced in the manner 
just illustrated. A flexible inextensible cord may be considered 
as made up of a great number of short rigid bodies jointed 
without friction, and hence may form part of a machine with- 
out vitiating the truth of (XYL). 

2(K. E.) n signifies the sum obtained by adding the amounts 
of kinetic energy {^dMv^ for each elementary mass) possessed 
by all the particles of all the rigid bodies at their final posi- 
tions ; 2(K. E.) , a similar sum at their initial positions. For 
example, the K. E. of a rigid body having a motion of transla- 
tion of velocity v, = %v 2 fdM= \ Mv* ; that of a rigid body 
having an angular velocity gd about a fixed axis Z, = \^Iz 
>(§ 139) ; while, if it has an angular velocity go about a gravity- 



150 MECHANICS OF ENGINEERING. 

axis Z, which has a velocity v z of translation at right angles to 
itself, the (K. E.) at this instant may be proved to be 

iMv z ' + ioo'I z , 

i.e., is the sum of the amounts due to the two motions sepa- 
rately. 

143. K. E. of Combined Rotation and Translation. — The last 
statement may be thus proved. Fig. 157. 
At a given instant the velocity of any dM is 
v, the diagonal formed on the velocity v z of 
~^^ji\ V * translation, and the rotary velocity oop rela- 
2H--^ ;/ \ tively to the moving gravity-axis Z (per- 

' \ pendicular to paper) (see §71), 

"A 



Fig. 158. ^; V* = V z + (™Pf ~ Z(™P)v Z COS <p \ 

hence we have K. E., at this instant, 
= fidMv* = iv z ydM + WfdMp* - cov z fdMp cos <?, 

but p cos cp = y, and fdMy = My = 0, since Z is a gravity- 
axis, 

.-. K E. = iMv z * + \*?I Z . Q. E. D. 

It is interesting to notice that the K. E. due to rotation, viz.>, 
\c&?I z = \M{ooTcf, is the same as if the whole mass were con- 
centrated iii a point, line, or thin shell, at a distance k, the 
radius of gyration, from the axis. 

144. Example of a Machine in Operation. — Fig. 159. Con- 
sider the four consecutive moving masses, J/ 7 , M", M"' ', and 
M' lv (being the piston ; connecting-rod ; fly-wheel, crank, drum,, 
and chain ; and weight on inclined plane) as free, collectively. 
Let us apply eq. (XVI.), the initial and final positions being 
taken when the crank-pin is at its dead-points o and n ; i.e., we 
deal with the progress of the pieces made while the crank-pin 
describes its upper semicircle. Remembering that the mutual 
actions between any two of these four masses can be left out 
of account (except friction), the only forces to be put in are 
the actions of other bodies on each one of these four, and are. 



WORK, ENERGY, AND POWER. 151 

shown in the figure. The only mutual friction considered will 
be at the crank-pin, and if this as an average — F\ the work 
done on it between o and n = F"7tr", where r" = radius of 
crank-pin. The work done by P x the effective steam- pressure 
(let it be constant) during this period is = PJJ ; that done in 
overcoming^, the friction between piston and cylinder, = FJ! ; 
that done upon the weight G"of connecting-rod is cancelled by 
the work done by it in the descent following ; the work done 




Fig. 159. 

upon G iv 9 = G iv 7ta sin /?, where a = radius of drum ; that 
upon the friction F A , = Fjta. The pressures iV 7 , J¥ iv , and 
iV 7 ", and weights G' and G" ; , are neutral, i.e., do no work either 
positive or negative. Hence the left-hand member of (XVI.) 
becomes, between o and n, 

P X V - FJ/ - F'nr" - G^na sin ft - Fjza, . . (1) 

provided the respective distances are actually described by 
these forces, i.e., if the masses have sufficient initial kinetic 
energy to carry the crank-pin beyond the point of minimum 
velocity, with the aid of the working force P x , whose effect is 
small up to that instant. 

As for the total initial kinetic energy, i.e., 2(K. E.) , let us 
express it in terms of the velocity of crank-pin at o, viz., V . 
The (K.E.) of M' is nothing ; that of M", which at this in- 
stant is rotating about its right extremity {fixed for the instant) 
with angular velocity go" = V -i- I", is \Go ,n I^ n \ that of M'" 
= %<*/"*I c '", in which a/" = V -r- r ; that of M iv (translation) 
= %M iv v Q lv % in which v iv = {a-~r) V . ^(K. E.) n is expressed 



152 



MECHANICS OF ENGINEERING. 



in a corresponding manner witli Y n (final velocity of crank-pin) 
instead of V . Hence the right-hand member of (XYI.) will 
give (putting the radius of gyration of M" about 0" = k", 
and that of M" about C = h) 



a v: - v:)[_m^ + jr"£ +^ i >]. . 



(2) 



By writing (1)=(2), we have an equation of condition, capa- 
ble of solution for any one unknown quantity, to be satisfied 
for the extent of motion considered. It is understood that the 
chain is always taut, and that its weight and mass are neg- 
lected. 

145. Numerical Case of the Foregoing. — (Foot-pound-second 
system of units for space, force, and time; this requires g 

= 32.2.) 

Suppose the following data : 



Feet. 


Lbs. 


Lbs. 


Mass Units. 


V = 2.0 
I" = 4.0 
a = 1.5 
r = 1.0 
Jc = 1.8 
7c" = 3.3 
r" = 0.1 


Pi = 

Tp 

F" (av'ge) = 
F* = 


6000 
200 
400 
300 


O' = 60 
G" = 50 
Q>" = 400 
G^ = 3220 


(and .".) 

M' = 1.86 
M" = 1.55 
M"> = 12.4 
Jf iv = 100.0 


Also let V = 4 


ft. per sec.; fJ= 30° 



Denote (1) by IF" and the large bracket in (2) by M (this by 
some is called the total mass " reduced" to the crank-pin). 
Putting (1) = (2) we have, solving for the unknown V n , 



/2W 



(3) 



For above values, 

W = 12,000 - 400 - 125.7 - 7590.0 - 1417.3 
= 2467 foot-pounds ; 
while M = 0.5 + 40.3 + 225.0 = 265.8 mass-units ; 
whence* 

V n = 1/18.56 + 16= VslM = 5.88 ft. per second. 



WORK, ENERGY, AND POWER. 153 

As to whether the crank-pin actually reaches the dead-point 
7i, requires separate investigations to see whether Y becomes 
zero or negative between o and n (a negative value is inad- 
missible, since a reversal of direction implies a different 
value for W), i.e., whether the proposed extent of motion is 
realized ; and these are made by assigning some other inter- 
mediate position m, as a final one, and computing V m , remem- 
bering that when m is not a dead-point the (K. E.) m of M' is not 
zero, and must be expressed in terms of V m , and that the 
{K. E.) m of the connecting-rod J/ 7 ' must be obtained from § 143. 

146. Regulation of Machines. — As already illustrated in 
several examples (§ 121), a fly-wheel of sufficient weight and 
radius may prevent too great fluctuation of speed in a single 
stroke of an engine ; but to prevent a permanent change, which 
must occur if the work of the working force or forces (such as 
the steam-pressure on a piston, or water-impulse in a turbine) 
•exceeds for several successive strokes or revolutions the work 
required to overcome resistances (such as friction, gravity, re- 
sistance at the teeth of saws, etc., etc.) through their respective 
spaces, automatic governors are employed to diminish the 
working force, or the distance through which it acts per stroke, 
until the normal speed is restored ; or vice versa, if the speed 
slackens, as when new resistances are temporarily brought into 
play. Hence when several successive periods, strokes (or other 
cycle), are considered, the kinetic energy of the moving parts 
will disappear from eq. (XVI.), leaving-it in this form : 

work of working-forces = work done upon resistances. 

147. Power of Motors. — In a mill where the same number of 
machines are run continuously at a constant speed proper for 
their work, turning out per hour the same number of barrels 
of flour, feet of lumber, or other commodity, the motor (e.g., 
a steam-engine, or turbine) works at a constant rate, i.e., de- 
velops a definite horse-power (H.P.), which is thus found in 
the case of steam-engines (double-acting) : 



154 MECHANICS OF ENGINEERING. 

H.P. = total mean effective ) / distance in feet ) 

steam-pressure on I X ] travelled by pis- > -f- 550 r 
piston in lbs. ) ( ton per second. ) 

i.e., the work (in ft.-lbs) done per second by the working force 
divided by 550 (see § 132). The total effective pressure at any 
instant is the excess of the forward over the back-pressure,, 
and by its mean value (since steam is usually used expansively) 
is meant such a value P' as, multiplied by the length of stroke 
Z, shall give 



P'l =fpdx, 



where P is the variable effective pressure and dx an element 
of its path. If u is the number of strokes per second, we may 
also write (foot-pound- second system) 

H.P. = P'lu -- 550 = \j^Pdx~\u -r- 550. (XVII.) 

Very often the number of revolutions^?^ minute, m, of the 
crank is given, and then 

H.P. = P' (lbs.) X 2Z (feet) X m -r- 33,000. 

If F = area of piston we may also write P' — Fp', where p* 
is the mean effective steam-pressure per unit of area. Evi- 
dently, to obtain P' in lbs., we multiply J^in sq. in. by p' in 
lbs. per sq. in., or F in sq. ft. by p f in lbs. per sq. foot ; the 
former is customary, p' in practice is obtained by measure- 
ments and computations from " indicator-cards" (see § 135, in 
which (P 9 — P x ) corresponds to P of this section) ; or P'l, i.e.,. 

/ Pdx, may be computed theoretically as in § 59, Problem 4. 

The power as thus found is expended in overcoming the 
friction of all moving parts (which is sometimes a large item), 
and the resistances peculiar to the kind of work done by the ma- 
chines. The work periodically stored in the increased kinetic 
energy of the moving masses is restored as they periodically 
resume their minimum velocities. 



WORK, ENERGY, AND POWER. 155 

148. Potential Energy. — There are other ways in which work 
or energy is stored and then restored, as follows : 

First. In raising a weight G through a height h, an amount 
of work = Gh is done upon G, as a resistance, and if at any 
subsequent time the weight is allowed to descend through the 
same vertical distance h (the form of path is of no account), G y 
now a working force, does the work Gh, and thus in aiding the 
motor repays, or restores, the Gh expended by the motor in 
raising it. If h is the vertical height through which the centre 
of gravity rises and sinks periodically in the motion of the 
machine, the force G may be left out of account in reckoning 
the expenditure of the motor's work, and the body when at its 
highest point is said to possess an amount Gh of potential 
energy, i.e., energy of position, since it is capable of doing the 
work Gh in sinking through its vertical range of motion. 

Second. So far, all bodies considered have been by express 
stipulation rigid, i.e., incapable of changing shape. To see 
the effect of a lack of rigidity as affecting the principle of 
work and energy in machines, 
take the simple case in Fig. 160. 
A helical spring at a given in- 
stant is acted on at each end by 
a force P in an axial direction 
(they are equal, supposing the Fig. 160. 

mass of the spring small). As the machine operates of which 
it is a member, it moves to a new consecutive position B, 
suffering a further elongation dX in its length (if P is increas- 
ing). P on the right, a working force, does the work Pdx'; 
how is this expended 1 P on the left has the work Pdx done 
upon it, and the mass is too small to absorb kinetic energy or 
to bring its weight into consideration. The remainder, Pdx' 
— Pdx = PdX, is expended in stretching the spring an addi- 
tional amount dX, and is capable of restoration if the spring 
retains its elasticity. Hence the work done in changing the 
form of bodies if they are elastic is said to be stored in the 
form of potential energy. That is, in the operation of ma- 
chines, the name potential energy is also given to the energy 




156 MECHANICS OF ENGINEERING. 

stored and restored periodically in the changing and regaining 
of form of elastic bodies. 

149. Other Forms of Energy. — Numerous experiments with 
■various kinds of apparatus have proved that for eveiy 772 
(about) ft.-lbs. of work spent in overcoming friction, one British 
unit of heat is produced (viz., the quantity of heat necessary to 
raise the temperature of one pound of w T ater from 32° to 33° 
Fahrenheit); while from converse experiments, in which the 
amount of heat used in operating a steam-engine was all carefully 
estimated, the disappearance of a certain portion of it could only 
be accounted for by assuming that it had been converted into 
work at the same rate of (about) 772 ft.-lbs. of work to each 
unit of heat (or 425 kilogrammetres to each French unit of 
heat). This number 772, or 425, according to the system of 
units employed, is called the Mechanical Equivalent of Heat, 
first discovered by Joule and confirmed by Him. 

Heat then is energy, and is supposed to be of the kinetic 
form due to the rapid motion or vibration of the molecules of 
a substance. A similar agitation among the molecules of the 
(hypothetical) ether diffused through space is supposed to pro- 
duce the phenomena of light, electricity, and magnetism. 
Chemical action being also considered a method of transform- 
ing energy (its possible future occurrence as in the case of coal 
and oxygen being called potential energy), the well-known 
•doctrine of the Conservation of Energy, in accordance with 
which energy is indestructible, and the doing of work is simply 
the conversion of one or more kinds of energy into equivalent 
amounts of others, is now one of the accepted hypotheses of 
physics. 

"Work consumed in friction, though practically lost, still re- 
mains in the universe as heat, electricity, or some other subtile 
form of energy. 

150. Power Required for Individual Machines. Dynamome- 
ters of Transmission. — If a machine is driven by an endless 
belt from the main-shaft, A, Fig. 161, being the driving-pulley 



WORK, ENERGY, AND POWER 



157 



on the machine, the working force which drives the machine, 
in other words the " grip" with which the 
belt takes hold of the pulley tangentially ; 
= P — P\ P and P' being the tensions 
in the " driving" and " following" sides of 
the belt respectively. The belt is supposed 
not to slip on the pulley. If v is the ve- 
locity of the pulley -circumference, the 
work expended on the machine per second, i.e., the power, 




Fig. 161. 



Z = (P- P')v. 



is 



(1) 



To measure the force (P — P'\ an apparatus called a Dy- 
namometer of Transmission maybe placed between the main 
shaft and the machine, and the belt made to pass through it in 
such a way as to measure the tensions P and P\ or princi- 
pally their difference, without meeting any resistance in so do- 
ing ; that is, the power is transmitted, not absorbed, by the 
apparatus. One invention for this purpose (mentioned in the 
Journal of the Franklin Institute some years ago) is shown 

{in principle) in Fig. 162. A ver- 
tical plate carrying four pulleys and 
a scale-pan is first balanced on the 
pivot C. The belt being then ad- 
justed, as shown, and the power 
turned on, a sufficient weight G is 
placed in the scale-pan to balance 
the plate again, for whose equilib- 
rium we must have Go = Pa — P'a, since the P and P' on 
the right are purposely given no leverage about C. The ve- 
locity of belt, v, is obtained by a simple counting device.. 
Hence (P — P') and v become known, and .*. L from (1). 

Many other forms of transmission-dynamometers are in use, 
some applicable whether the machine is driven by belting or 
gearing from the main shaft. Emerson's Hydrodynamics de- 
scribes his own invention on p. 283, and gives results of meas- 
urements with it ; e.g., at Lowell, Mass., the power required 
to drive 112 looms, weaving 36-inch sheetings, No. 20 yarn,. 




158 MECHANICS OF ENGINEERING. 

60 threads to the inch, speed 130 picks to the minute, was 
found to be 16 H.P., i.e., \ H.P. to each loom (p. 335). 

151. Dynamometers of Absorption. — These are so named 
since they furnish in themselves the resistance (friction or a 
weight) in the overcoming (or raising) of which the power is 
expended or absorbed. Of these the Prony Friction Brake 
is the most common, and is used for measuring the power 
developed by a given motor (e.g., a steam-engine or turbine) 
not absorbed in the friction of the motor itself. Fig. 163 




Fig. 163. 



shows one fitted to a vertical pulley driven by the motor. By 
tightening the bolt B, the velocity <v of pulley-rim may be 
made constant at any desired value (within certain limits) by 
the consequent friction, v is measured by a counting appara- 
tus, while the friction (or tangential components of action be- 
tween pulley and brake), = F, becomes known by noting the 
weight G which must be placed in the scale-pan to balance the 
arm between the checks ; then 

Fa=Gb, (1) 

for the equilibrium of the brake (supposing the weight of 
brake and scale-pan previously balanced on C) and the work 
done per unit of time, or power, is 

Z = Fv. (2) 

A " dash-pot " is frequently connected with the arm to prevent 
sudden oscillations. In case the pulley is horizontal, a bell- 
crank lever is added between the arm and the scale-pan, and 
then eq. (1) will contain two additional lever-arms. 



WORK, ENERGY, AND POWER 



159 



< — 








l — 





1 


Po 


R 


P 2 


P 3 


P* 


P 5 |^ 














Pe 



152. The Indicator, used with steam and other fluid engines, 
Is a special kind of dynamometer in which the automatic mo- 
tion of a pencil describes a curve 
on paper whose ordinates are 
proportional to the fluid pres- 
sures exerted in the cylinder at 
successive points of the stroke. 
Thus, Fig. 164, the back-pres- 
sure being constant and = P b , 
the ordinates P , P x , etc., represent the effective pressures at 
equally spaced points of division. The mean effective pressure 
P' (see § 147) is, for this figure, by Simpson's Kule (six equal 
spaces), 

P' = -frl?. + KP, + P, + P>) + HP, + P<) + P.l 

This gives a near approximation. The power is now found by 
§147. 



Fig. 164. 



153. The theory of Atwood's Machine is most directly ex- 
pressed by the principle of work and energy ; i.e., by eq. 
(XVL), §142. Fig. 165. The parts 
considered free, collectively, are the 
rigid bodies P, Q, G, and four friction- 
wheels like G x \ and the flexible cord, 
which does not slip on the upper pul- 
ley. There is no slipping at D, hence 
no sliding friction there. The actions 
of external bodies on these eight consist 
of the working force P, the resistances 
Q and the four P's (at bearings of fric- 
tion-wheel axles); all others (6r, 46r x , 
and the four i?'s) are neutral. Since there is no rubbing be- 
tween any two of the eight bodies, no mutual actions whatever 
will enter the equation. Let P > Q, and /and I x be the mo- 
ments of inertia of G and G„ respectively, about their respec- 
tive axes of figure. Let the apparatus start from rest, then 
when P has descended through any vertical distance s, and ac- 




160 MECHANICS OF ENGINEERING. 

quired the velocity v, Q has been drawn up an equal distance- 
and acquired the same velocity, while the pulley G has ac- 
quired an angular velocity go = v -f- a, each friction-pulley an 
angular velocity od 1 — (r : a)v -r- a x . As to the forces, P has 
done the work Ps, Q has had the work Qs done upon it, while 
each F has been overcome through the space (r x : af)(r : a)s ; 
all the other forces are neutral. Hence, from eq. (XVL), § 142 
(see also § 139), we have 

p s -Q s - 4F- 1 . -s 

Lg ' g J2 ' 2 a '2 a? a? x 

Evidently v — Vs X constant, i.e., the motion of P and Q is 
uniformly accelerated. If, after the observed space s has been 
described, P is suddenly diminished to such a value P' that 
the motion continues with a constant velocity = v, we shall 
have, for any further space s' 9 

P's' -Qs' -±F^.-s' = 0, 

from which jPcan be obtained (nearly) ; while if if be the ob- 
served time of describing s', v == s' -f 2 if becomes known. 
Also we may write 1= {G ~- g)¥ and I x = (G x -f- ^O^ 2 , and 
thus finally compute the acceleration of gravity, g, from our 
first equation above. 

154. Boat-Rowing. — -Fig. 166. During the stroke proper, 
let P = mean pressure on one oar-handle ; hence the pressures 
on the foot-rest are 2P, resistances. Let Jf=massof boat 
and load, v and v n its velocities at beginning and end of stroke. 
P l = pressures between oar- blade and water. R = mean re- 
sistance of water to the boat's passage at this (mean) speed. 
These are the only (horizontal) forces to be considered as act- 
ing on the boat and two oars, considered free collectively. 
During the stroke the boat describes the space s 3 = CD, the- 
oar-handle the space s a = AB, while the oar-blade slips back- 



WORK, ENERGY, AND POWER. 



161 



ward through the small space (the " slip") == s 1 (average). 
Hence by eq. (XVI.), § 142, 

2Ps, - 2Ps 3 -It s, - 2P A = \M{< - <) ; 
i.e., 2P(s-s 3 )=2PxA£=2Ps =Ps 3 +2P lSl + iM(v n *-v;); 

or, in words, the product of the oar-handle pressures into the 
distance described by them measured on the boat, i.e., the work 
done by these pressures relatively to the boat, is entirely ac- 
counted for in the work of slip and of liquid-resistance, and in- 




Fia 166. 

creasing the kinetic energy of the mass. (The useless work 
due to slip is inevitable in all paddle or screw propulsion, as 
well as a certain amount lost in machine-friction, not considered 
in the present problem.) During the " recover" the velocity 
decreases again to v . 

155. Examples. — 1. "What work is done on a level track, in 
bringing up the velocity of a train weighing 200 tons, from 
zero to 30 miles per hour, if the total frictional resistance (at 
any velocity, say) is 10 lbs. per ton, and if the change of speed 
is accomplished in a length of 3000 feet % 

{Foot-ton-second system.) 30 miles per hour = 44 ft. per 
sec. The mass 

= 200 -f- 32.2 = 6.2 ; 
•\ the change in kinetic energy, 

(= iMv* -iMx 3 ), 
= £(6.2) X 44 2 = 6001.6 ft.-tons. 
11 



162 MECHANICS OF ENGINEERING. 

The work done in overcoming friction = Es, i.e., 

= 200 X 10 X 3000 = 6,000,000 ft.-lbs. = 3000 ft.-tons ; 
.-. total work = 6001.6 + 3000 = 9001.6 ft.-tons. 

(If the track were an up-grade, 1 in 100 say, the item of 
500 X 30 = 6000 ft.-tons would be added.) 

Example 2. — Required the rate of work, or power, in Ex- 
ample 1. The power is variable, depending on the velocity of 
the train at any instant. Assume the motion to be uniformly 
accelerated, then the working force is constant ; call it P. 
The acceleration (§ 56) will be^=v a -^2s=1936-=-6000=0.322 
ft. per sq. sec; and since P — F ' = Mp, we have 

P = 1 ton + (200 -T- 32.2) X 0.322 = 3 tons, 

which is 6000 -~ 200 = 30 lbs. per ton of train, of which 20 is 
due to its inertia, since when the speed becomes uniform the 
work of the engine is expended on friction alone. 

Hence when the velocity is 44 ft. per sec, the engine is 
working at the rate of Pv = 264,000 ft.-lbs. per sec, i.e., at the 
rate of 480 H. P.; 

At J of 3000 ft. from the start, at the rate of 240 H. P., half 
as much ; 

At a uniform speed of 30 miles an hour the power would be 
simply 1 X 44 = 44 ft. -tons per sec. = 160 H. P. 

Example 3. — The resistance offered by still water to the 
passage of a certain steamer at 10 knots an hour is 15,000 lbs. 
What power must be developed by its engines, at this uniform 
speed, considering no loss in "slip" nor in friction of ma- 
chinery ? Ans. 461 H. P. 

Example 4. — Same as 3, except that the speed is to be 15 
knots (i.e., nautical miles ; each = 6086 feet) an hour, assum- 
ing that the resistances are as the square of the speed (approxi- 
mately true). Ans. 1556 H. P. 

Example 5. — Same as 3, except that 12$ of the power is ab- 
sorbed in the " slip" (i.e., in pushing aside and backwards the 
water acted on by the screw or paddle), and %% in friction of 
machinery. Ans. 576 H. P. 

Example 6. — In Example 3, if the crank-shaft makes 60 



WORK, ENERGY, AND POWER. 163 

revolutions per minute, the crank-pin describing a circle of 18 
inches radius, required the average value of the tangential 
component of the thrust (or pull) of the connecting-rod against 
the crank-pin. Ans. 26890 lbs. 

Example 7. — A solid sphere of cast-iron is rolling up an in- 
cline of 30°, and at a certain instant its centre has a velocity of 
36 inches per second. Neglecting friction of all kinds, how 
much further will the ball mount the incline (see § 143) % 

Ans. 0.390 ft. 

Example 8. — In Fig. 163, with h = 4 ft. and a = 16 inches, 
it is found in one experiment that the friction which keeps the 
speed of the pulley at 120 revolutions per minute is balanced 
by a weight G = 160 lbs. Required the power thus measured. 

Although in Examples 1 to 6 the steam cylinder is itself in 
motion, the work" per stroke is still = mean effective steam- 
pressure on piston X length of stroke, for this is the final form 
to which the separate amounts of work done by, or upon, the 
two cylinder heads and the two sides of the piston will re- 
duce, when added algebraically. See § 154. Ans. 14.6 H. P. 



164 MECHANICS OF ENGINEERING. 



CHAPTER VII. 

FRICTION. 

156. Sliding Friction. — When the surfaces of contact of two 

bodies are perfectly smooth, the direction of the pressure or pair 

of forces between them is normal to these surfaces, i.e., to their 

> tangent-plane ; but when they are rough, and 

Y"? I moving one on the other, the forces or ac- 

pV"""4n tions between them incline away from the 

;\ ; l ^ p^ normal, each on the side opposite to the di- 

P/S/A/fa m '/'//J^/////yp rect i° n °f the (relative) motion of the body 

mm^mm on whicll it acts> ThuSj Fig> 167? a block 

fig. 167. whose weight is G, is drawn on a rough 

horizontal table by a horizontal cord, the tension in which is 
P. On account of the roughness of one or both bodies the ac- 
tion of the table upon the block is a force P x , inclined to the 
normal (which is vertical in this case) at an angle = <p away 
from the direction of the relative velocity v. This angle <p is 
called the angle of friction, while the tangential component of 
P x is called the friction = F. The normal component N y 
which in this case is equal and opposite to G the weight of the 
body, is called the normal pressure. 

Obviously F — iTtan <p, and denoting tan <p bjf, we have 

F=fN. (1) 

/"is called the coefficient of friction, and may also be defined 
as the ratio of the friction i^to the normal pressure N which 
produces it. 



FRICTION. 165 

In Fig. 167, if the motion is accelerated (ace. =p), we have 
<eq. (IV.), § 55) P - F = Mp ; if uniform, P-F=Q; from 
which equations (see also (1))/ may be computed. In the 
latter case /* may be found to be different with different veloci- 
ties (the surfaces retaining the same character of course), and 
then a uniformly accelerated motion is impossible unless P 
— F were constant. 

As for the lower block or table, forces the equals and op- 
-posites of i\^andi^(or a single force equal and opposite to P x ) 
are comprised in the system of forces acting upon it. 

As to whether F is a working force or a resistance, when 
either of the two bodies is considered free, depends on the cir- 
cumstances of its motion. For example, in friction-gearing 
the tangential action between the two pulleys is a resistance 
for one, a working force for the other. 

If the force P, Fig. 167, is just sufficient to start the body, 
•or is just on the point of starting it (this will be called impending 
motion), Fis called the frictio?i of rest. If the body is at rest 
and P is not sufficient to start it, the tangential component will 
then be < the friction of rest, viz., just = P. As P increases, 
this component continually equals it in value, and P 1 acquires 
a direction more and more inclined from the normal, until the 
instant of impending motion, when the tangential component 
z=f]V= the friction of rest. When motion is once in prog- 
ress, the friction, called then the friction of motion, = fJV, 
in which y is not necessarily the same as in the friction of rest. 

157. Laws of Sliding Friction. — Experiment has demon- 
strated the following relations approximately, for two given 
rubbing surfaces : (See § 175.) 

(1) The coefficient, f is independent of the normal pressure 
IT. 

(2) The coefficient,/", for friction of motion, is the same at 
all velocities. 

(3) The coefficient, f for friction of rest (i.e., impending 
motion) is usually greater than that for friction of motion 
(probably on account of adhesion). 



166 



MECHANICS OF ENGINEERING. 



, (4) The coefficient, y, is independent of the extent of rub- 
bing surface. 

(5) The interposition of an unguent (such as oil, lard, tallow, 
etc.) diminishes the friction very considerably. 

158. Experiments on Sliding Friction. — These may be made 
with simple apparatus. If a block of weight = G, Fig. 168, 
be placed on an inclined plane of uniformly rough surface, 
and the latter be gradually more and more inclined from the 
horizontal until the block begins to move, the value of /? at 





Fig. 168. 



Fig. 169. 



this instant = <p, and tan q?=f= coefficient of friction of 
rest. For from 2X =0 we have F, i.e., fN, = G sin /?;. 
from 2Y= 0, JV '== G cos /3 ; whence tan /3 = f, .*. fi must 

= 9> 

Suppose fi so great that the motion is accelerated, the body 

starting from rest at o, Fig. 169. It will be found that the 
distance x varies as the square of the time, hence (§ 56) the 
motion is uniformly accelerated (along the axis X). (Notice 
in the figure that G is no longer equal and opposite to P^ the 
resultant of iTand F, as in Fig. 168.) 

2Y= 0, which gives JV — G cos /? = ; 

2X = Mp, which gives G sin fi—fN= (G ~ g)p ; 

while (from § 56) p=2se-h f. 

Hence, by elimination, x and the corresponding time t having 
been observed, we have for the coefficient of friction of motion^ 



/=tany3 



2a? 



gtf cos /?' 



FRICTION. 167 

In view' of (3), § 157, it is evident that if a value fi m has been 
found experimentally fox- ft such that the block, once started by 
hand, preserves a uniform motion down the plane, then, since 
tan fi m =f for friction of motion, /3 m may be less than the /3 
in Fig. 168, for friction of rest. 

159. Another apparatus consists of a horizontal plane, a pul- 
ley, cord, and two weights, as shown in Fig. 59. The masses 
of the cord and pulley being small and hence neglected, the 
analysis of the problem when G is so large as to cause an ac- 
celerated motion is the same as in that example [(2) in § 57], 
except in Fig. 60, where the frictional resistance yiT should be 
put in pointing toward the left. i\T still = G iy and .*. 

0-/8, = (<?,-*-&>; (1) 

while for the other free body in Fig. 61 we have, as before, 

G-S=(G + g)p (2) 

From (1) and (2), S the cord-tension can be eliminated, and 
solving for p, writing it equal to 2s -7- tf, s and t being the ob- 
served distance described (from rest) and corresponding time, 
we have finally for friction of motion 

G_ G + G, fe ... 

J ~ G, G\ • gf ( 3 > 

If G, Fig. 59, is made just sufficient to start the block, or 
sledge, G^ we have for the friction of rest 

/=! (*) 

160. Results of Experiments on Sliding Friction. — Professor 
Thurston in his article on Friction (which the student will do 
well to read) in Johnson's Cyclopaedia gives the following 
epitome of results from General Morin's experiments (made 
for the French Government in 1833) : 



168 



MECHANICS OF ENGINEERING. 
TABLE FOR FRICTION OF MOTION. 



No. 


Surfaces. 


Unguent. 


Angle <£. 


/ = tan <£. 


1 


Wood on wood. 


None. 


14° to 26£° 


0.25 to 0.50 


2 


Wood on wood. 


Soap. 


2° to lli° 


0.04 to 0.20 


3 


Metal on wood. 


None. 


26i° to 31i° 


0.50 to 0.60 


4 


Metal on wood. 


Water. 


15° to 20° 


0.25 to 0.35 


5 


Metal on wood. 


Soap. 


nr 


0.20 


6 


Leather on metal. 


None. 


29£° 


0.56 


7 


Leather on metal. 


Greased. 


13° 


0.23 


8 


Leather on metal. 


Water. 


20° 


0.36 


9 


Leather on metal. 


Oil. 


8|° 


0.15 


10 


Smoothest and best 










lubricated surfaces. 




lf°to 2° 


0.03 to 0.036 









For friction of rest, about 4:0% may be added to the coeffi- 
cients in the above table. 

In dealing with the stone blocks of an arch-ring, cp is com- 
monly taken = 30°, i.e., f = tan 30° = 0.58 as a low safe 
value ; it is considered that if the direction of pressure between 
two stones makes an angle > 30° with the normal to the joint 
(see § 161) slipping may take place (the adhesion of cement 
being neglected). 

General Morin states that for a sledge on dry ground f = 
about 0.66. 

"Weisbach gives for metal on metal, dry (E. R. brakes for 
example),/ = from 0.15 to 0.21. Trautwine's Pocket-Book 
gives values of f for numerous cases of friction. 

161. Cone of Friction.— Fig. 170. Let A and B be two 

rough blocks, of which B is immovable, and P the resultant 
of all the forces acting on A, except the pres- 
sure from B. B can furnish any required 
normal pressure N to balance P cos /?, but 
the limit of its tangential resistance is fN. 
So long then as fi is < cp the angle of fric- 
tion, or in other words, so long as the line of 
fig. 170. action of P is within the " cone of friction" 

generated by revolving OC about OJV, the block A will not 




FRICTION. 



169 



slip on B, and the tangential resistance of B is simply P sin 
/3 ; but if j3 is > g>, this tangential resistance being onlyfJY 
-and < P sin /?, A will begin to slip, with an acceleration. 

162. Problems in Sliding Friction. — In the following prob- 
lems/* is supposed known at points where rubbing occurs, or 
is impending. As to the pressure iV to which the friction is 
due, it is generally to be considered unknown until determined 
by the conditions of the problem. Sometimes it may be an 
advantage to deal with the single unknown force P (resultant 
of iTand/l/^) acting in a line making the known angle 9? with 
the normal (on the side away from the motion). 

Problem 1. — Required the value of the weight P, Fig. 171, 
the slightest addition to which will cause motion of the hori- 
zontal rod OB, resting on rough planes at 45°. The weight 
G of the rod may be applied at the 
middle. Consider the rod free ; at 
each point of contact there is an un- 
known JV and a friction due to it 
fN\ the tension in the cord will be 
= P, since there is no acceleration 
and no friction at pulley. Notice 
the direction of the frictions, both opposing the impending 
motion. [The student should not rush to the conclusion that 
N and N' 1 are equal, and are the same as would be produced by 
the components of G if the latter were transferred to A and 
resolved along AO and AB ; but should await the legitimate 
results deduced by algebra, from the equations of condition 
for the equilibrium of a system of forces in a plane. Few 
problems in Mechanics are so simple as to admit of an imme- 
diate mental solution on inspection ; and guess-work should be 
carefully avoided.] 

Taking an origin and two axes as in figure, we have (eqs. 
(2), § 36), denoting the sine of 45° by m, 

2X f^ + mG-JV -P = 0;. . (1) 

2Y iT x + /^- mG = 0;. . (2) 

2(Pa) fNa + Na- Gl = 0. . . (3) 




Fig. 171. 



170 



MECHANICS OF ENGINEERING. 



The three unknowns P, JV, and iV^ can now be found. 
Divide (3) by a, remembering that b : a = m, and solve for 
JT; substitute it in (2) and iV^ also becomes known ; while P 
is then found from (1) and is 

2 3 fG_/V2 

Problem 2. — Fig. 172. A rod, centre of gravity at middle, 
leans against a rough wall, and rests on an equally rough floor; 
how small may the angle a become before it 
slips ? Let a = the half-length. The figure 
p shows the rod free, and following the sugges- 
tion of § 162, a single unknown force P 1 
P making a known angle cp (whose tan =f) 
d with the normal BE, is put in at B, leaning 
away from the direction of the impending 
p motion, instead of an JV and fN\ similarly 
fig. 172. P^ acts at 0. The present system consisting 

of but three forces, the most direct method of finding «, with- 
out introducing the other two unknowns P 1 and P u at all, is 
to use the principle that if three forces balance, their lines 
of action must intersect in a point. That is, P^ must inter- 
sect the vertical containing 6r, the weight, in the same point 
as P l5 viz., A. 

Now EA, and also PC, = a cos a, 

.'. ED = a cos a cot cp and AB == a cos a tan cp* 
But DF, which = 2a sin a, = DE — AB \ 

.*. 2a sin a = a cos a [cot cp — tan <p~\. . . (1) 

Dividing by cos <*, and noting that tan cp =f= 1 -f- cot <p f 
we obtain for the required value of a 




tan a = 



i \-r 



f 



and finally, tan a = cot 2cp, 



after some trigonometrical reduction. That is, a is the com- 
plement of double the angle of friction. 



FRICTION. 



171 



Problem 3. — Fig. 173. Given the resistance Q, acting- 
parallel to the fixed guide C, the angle a, and the (equal) co- 
efficients of friction at the rubbing surfaces, required the 





Fig. 173. 



Fig. 174. 



amount of the horizontal force P, at the head of the block A 
(or wedge), to overcome Q and the frictions. D is fixed, and 
ab is perpendicular to cd. Here we have four unknowns, viz.> 
P, and the three pressures iV, N^ and N» between the blocks. 
Consider A and B as free bodies, separately (see Fig. 174), re- 
membering Newton's law of action and reaction. The full 
values (e.g.,fJV) of the frictions are put in, since we suppose 
a slow uniform motion taking place. 
For A, 2X= and 2Y = give 

i\T — iTcos a +yi\ r sin a — P sin a = ; . 

fN x -\-JV&in a -\-fj¥ cos a — P cos a = 0. . 

For ^, ^Xand JSZgive 

e--ar,+/y. = 0;....(8) and ^ -fJST x = 0. 

Solve (4) for iV^ and substitute in (3), whence 

*&-/*)= Q (5) 

Solve (2) for N, substitute the result in (1), as also the value 
of N 1 from (5), and the resulting equation contains but one un- 
known, P. Solving for P, putting for brevity 

ycos a-\- sin a = m and cos a — ysin a = n, 

(m+fn)Q 



(1) 

(2) 



we have P = 



(n . cos a -\- m . sin a)(l —f 2 )" 



(6) 



172 MECHANICS OF ENGINEERING. 

Numerical Example of Problem 3. — If Q = 120 lbs., f 
= 0.20 {an abstract number, and .*. the same in any system of 
units), while a = 14°, whose sine = 0.240 and cosine ~ .970, 
then 

m = 0.2x.97 + 0.24 = 0.43 and n = .97 — .2X-24 = 0.92, 

whence P = 0.640 = 76.8 lbs. 

While the wedge moves 2 inches P does the work (or exerts 
an energy) of 2 X 76.80 = 153.6 in.-lbs. = 12.8 ft.-lbs. 

For a distance of 2 inches described by the wedge horizon- 
tally, the block B (and .'. the resistance Q) has been moved 
through a distance = 2 X sin 14° = 0.48 in, along the guide 
C, and hence the work of 120 X 0.48 = 57.6 in.-lbs. has been 
done upon Q. Therefore for the supposed portion of the 
motion 153.6 — 57.6 = 96.0 in.-lbs. of work has been lost in 
friction (converted into heat). 

It is noticeable in eq. (6), that if f should = 1.00, P = oc ; 
and that if a = 90°, P = Q, and there is no friction (the 
weights of the blocks have been neglected). 

Problem 4. Numerical. — With what minimum pressure 

P should the pulley A be held against B, which it drives by 

B n , x m " friction al gearing," to transmit 2 H.P.; 

"•-7 P 4 & ^ if a = 45°, f for impending (relative) 

a U / N ^ motion, i.e., for impending slipping = 

fig. 175. 0.40, and the velocity of the pulley-rim 

is 9 ft. per second ? 

The limit-value of the tangential " grip" 

T = 2fJSr= 2 X 0.40 X P sin 45°, 
2 H. P. = 2 X 550 = 1100 ft.-lbs. per second. 

Putting T X 9 ft. = 1100, we have 

2 X 0.40 X4XPX9= 1100 ; .-. P = 215 lbs. 

Problem 6.— A block of weight G lies on a rough plane, 
inclined an angle /? from the horizontal ; find the pull P, mak- 
ing an angle a with the first plane, which will maintain a uni- 
form motion up the plane. 



FRICTION. 173 

Problem 7. — Same as 6, except that the pull P is to permit 
a uniform motion dow?v the plane. 

Problem 8. — The thrust of a screw-propeller is 15 tons. 
The ring against which it is exerted has a mean radius of 8 
inches, the shaft makes one revolution per second, andy = 0.06. 
Required the H. P. lost in friction from this cause. 

Ana. 13.7 H. P. 

163. The Bent-Lever with Friction. Worn Bearing. — Fig. 
176. Neglect the weight of the lever, and suppose the plumb- 
er-block so worn that there is d,t .HE /^\ 

contact along one element only of / \\ ,//" / / P 

the shaft. Given the amount and gZ y \4- / / 

line of action of the resistance i?, / \ \ / / "*^* 

and the line of action of P, re- /\^^y^C-A-J[ 
quired the amount of the latter for / ^s. ;H -~^v\ 
impending slipping in the direction * R 
of the dotted arrow. As P grad- 
ually increases, the shaft of the 
lever (or gear-wheel) rolls on its fig. 176. 

bearing until the line of contact has reached some position A, 
when rolling ceases and slipping begins. To find A, and the 
value of P, note that the total action of the bearing upon the 
lever is some force P„ applied at A and making a known 
angle cp (f = tan cp) with the normal A C. P 1 must be equal 
and opposite to the resultant of the known R and the unknown 
P, and hence graphically (a graphic is much simpler here than 
an analytical solution) if we describe about (7 a circle of radius 
= r sin cp, r being the radius of shaft (or gudgeon), and draw 
a tangent to it from D, we determine PA as the line of action 
of P x . If DG is made = i?, to scale, and GF drawn parallel 
to D . . . P, P is determined, being = DE, while P x = DF. 

If the known force P is capable of acting as a working force,, 
by drawing the other tangent DB from D to the " friction- 
circle," we have P = PP, and P x = DK, for impending 
rotation in an opposite direction. 

If P and P are the tooth-pressures upon two spur-wheels, 
keyed upon the same shaft and nearly in the same plane, the. 



174 



MECHANICS OF ENGINEERING. 



[P x sin <p] 27tr. 



same constructions hold good, and for a continuous uniform 
motion, since the friction = P x sin tp, 
the work lost in friction 
per revolution, 

It is to be remarked, that without friction P x would pass 
through C, and that the moments of P and P would balance 
about C (for rest or uniform rotation) ; whereas with friction 
the j balance about the proper tangent-point of the friction- 
circle. 

Another way of stating this is as follows : Sp long as the 
resultant of P and P falls within the " dead-angle" PDA, 
motion is impossible in either direction. 

If the weight of the lever is considered, the resultant of it 
and the force P can be substituted for the latter in the fore- 
going. 

164. Bent-Lever with Friction. Triangular Bearing. — Like 

the preceding, the gudgeon is much exaggerated in the figure 

(177). For impending rotation in 
direction of the force P, the total 
actions at A x and A 2 must lie in 
known directions, making angles = <p 
with the respective normals, and in- 
clined away from the slipping. Join 
the intersections D and L. Since 
the resultant of P and P at D must 
act along PL to balance that of P x 
and P„ having given one force, say 
P, we easily find P = PP, while 
P x and P 2 = LM and LN respectively, LO having been made 
= PP, and the parallelogram completed. 

(If the direction of impending rotation is reversed, the change 
in the construction is obvious.) If P t = 0, the case reduces 
to that in Fig. 176 ; if the construction gives P 2 negative, the 
supposed contact at A, is not realized, and the angle A % CA X 
should be increased, or shifted, until P^ is positive. 

As before, P and P may be the tooth- pressures on two 




Fig. 177. 



FRICTION. 175 

spur-wheels nearly in the same plane and on the same shaft ; 
if so, then, for a uniform rotation, 
Work lost in fric. per re vol. = [P x sin cp -\- P s sin <p]%7tr. 

165. Axle-Friction. — The two foregoing articles are intro- 
ductory to the subject of axle -friction. When the bearing is 
new, or nearly so, the elements of the axle which are in contact 
with the bearing are infinite in number, thus giving an infinite 
number of unknown forces similar to P x and JP a of the last 
paragraph, each making an angle cp with its normal. Refined 
theories as to the law of distribution of these pressures are of 
little use, considering the uncertainties as to the value of 
f ( = tan (p) ; hence for practical purposes axle-friction may be 
written 

F=fB, 

in which f is a coefficient of axle-friction derivable from 
experiments with axles, and JR, the resultant pressure on the 
bearing. In some cases B, may be partly due to the tightness 
of the bolts with which the cap of the bearing is fastened. 

As before, the work lost in overcoming axle-friction per 
revolution is =fP27tr, in which r is the radius of the axle. 
f, like f is an abstract number. As in Fig. 176, a " friction- 
circle," of radius =fr, may be considered as subtending the 
" dead-angle." 

166. Experiments with Axle-Friction. — Prominent among 
recent experiments have been those 
of Professor Thurston (1872-73), 
who invented a special instrument 
for that purpose, shown (in princi- 
ple only) in Fig. 178. By means of 
an internal spring, the amount of 
whose compression is read on a scale, 
a weighted bar or pendulum is caused 
to exert pressure on a projecting axle 
from which it is suspended. The 
axle is made to rotate at any desired 
Telocity by some source of power, the axle-friction causing 




176 MECHANICS OF ENGINEERING. 

the pendulum to remain at rest at some angle of deviation 
from the vertical. The figure shows the pendulum free, the 
action of gravity upon it being G, that of the axle consisting 
of the two pressures, each = i?, and of the two frictions (each 
being F —f R\ due to them. Taking moments about 0, we 
have for equilibrium 

2fjRr = Gb, 

in which all the quantities except f are known or observed. 
The temperature of the bearing is also noted, with reference 
to its effect on the lubricant employed. Thus the instrument 
covers a wide range of relations. 

General Morin's experiments as interpreted by Weisbach 
give the following practical results : 

-> r 0.054 for well-sustained 

For iron axles, in iron or i lubrication; 

brass bearings '•' ] 0.07 to .08 for ordinary 

I lubrication. 

By " pressure per square inch on the bearing" is commonly 
meant the quotient of the total pressure in lbs. by the area in 
square inches obtained by multiplying the width of the axle by 
the length of bearing (this length is quite commonly four times 
the diameter) ; call it j?, and the velocity of rubbing in feet per 
minute, v. Then, according to Rankine, to prevent overheat- 
ing, we should have 

p(v + 20) < 44800 . . . (not homog.). 

Still, in marine-engine bearings pv alone often reaches 60,000, 
as also in some locomotives (Cotterill). Good practice keeps 
p within the limit of 800 (lbs. per sq. in.) for other metals 
than steel (Thurston), for which 1200 is sometimes allowed. 

With v = 200 (feet per min.) Professor Thurston found that 
for ordinary lubricants p should not exceed values ranging 
from 30 to 75 (lbs. per sq. in.). 

The product pv is obviously proportional to the power ex- 
pended in wearing the rubbing surfaces, per unit of area. 



FRICTION. 



177 



167. Friction-Wheels. — A single example of their use will 
be given, with some approximations to avoid complexity. Fig. 
179. G is the weight of a heavy wheel, P x is a known vertical 
resistance (tooth-pressure), and P an 
unknown vertical working force, 
whose value is to be determined to 
maintain a uniform rotation. The 
utility of the friction-wheels is also 
to be shown. The resultant of _P l5 
G, and P is a vertical force i?, pass- 
ing nearly through the centre C of 
the main axle which rolls on the four 
friction-wheels. B, resolved along 
CA and CB, produces (nearly) equal 
pressures, each being i\T = P -f- 2 cos <*, at the two axles of 
the friction- wheels, which rub against their fixed plumber- 
blocks. R = P -\- P i -f- 6r 1? and .*. contains the unknown P, 
but approximately = G -\- 2P^ i.e., is nearly the same (in this 
case) whether friction-wheels are employed or not. 

When G makes one revolution, the friction f'N at each axle 
C t is overcome through a distance = (?\ : a^) 27tr, and 

Work lost per revol. 

with 

friction-wheels, 




Fig. 179. 



= 2fJV i2*r = 
a 1 



a, cos a 



fP27tr. 



Whereas, if C revolved in a fixed bearing, 

=fP27rr. 



Work lost per revol. 

without 

friction-wheels, 



Apparently, then, there is a saving of work in the ratio r x \ 
a x cos a, but strictly the P is not quite the same in the two cases ; 
for with friction-wheels the force P is less than without, and P 
depends on P as well as on the known G and P x . By dimin- 
ishing the ratio r r : a 1} and the angle ar, the saving is increased. 
If a were so large that cos a < r 1 : # 1? there would be no saving, 
but the reverse. 

As to the value of P to maintain uniform rotation, we have 
12 



178 MECHANICS OF ENGINEERING. 

for equilibrium of moments about C, with friction-wheels (con- 
sidering the large wheel and axle free), 

Ph = P 1 h 1 + 2Tr, (1) 

In which T is the tangential action, or "grip," between one 
pair of friction-wheels and the axle C which rolls upon them. 
T would not equal fN unless slipping took place or were im- 
pending at E, but is known by considering a pair of friction- 
wheels free, when 2 (Pa) about C x gives 

1 J x J 2 cos a' 
which in (1) gives finally 

b TV 

P = ^P x -\ r -i—fR r (2) 

b ' ' a x cos ay b v ' 

Without friction-wheels, we would have 

P=\p,+fR\ (3) 

The last term in (2) is seen to be less than that in (3) (unless 
a is too large), in the same ratio as already found for the saving 
of work, supposing the i?'s equal. 

If P x were on the same side of C as P, it would be of an 
opposite direction, and the pressure P would be diminished. 
Again, if P were horizontal, P would not be vertical, and the 
friction-wheel axles would not bear equal pressures. Since P 
depends on P„ G, and the frictions, while the friction depends 
on P, and P on P x , G, and P, an exact analysis is quite 
complex, and is not warranted by its practical utility. 

Example. — If an empty vertical water-wheel weighs 25,000 
lbs., required the force P to be applied at its circumference to 
maintain a uniform motion, with a = 15 ft., and r = 5 inches. 
Here P x = 0, and P — G (nearly ; neglecting the influence of 
P on P), i.e., P = 25,000 lbs. 

First, without friction-wheels (adopting the foot-pound-sec- 
ond system of units), with/* = .07 (abstract number). From 
eq. (3) we have 

p = 0+0.07 X 25,000 X (A -*- 15 ) = ±8.6 lbs. 



FRICTION. 179 

The work lost in friction per revolution is 
fB%7tr = 0.07 X 25,000 X 2 X 3.14 X A = 4580 ft- 11 * 3 - 
/Secondly, with friction-wheels, in which r x : a x = \ and 
<$os a = 0.80 (i.e., a = 36°). From eq. (2) 

P = + | . -yi- X 48.6 = only 12.15 lbs., 

while the work lost per revolution 

= j. . ic> x 4580 = 1145 f t.-lbs. 

Of course with friction- wheels the wheel is not so steady as 
without. 

In this example the force P has been simply enough to 
overcome friction. In case the wheel is in actual use, P is the 
weight of water actually in the buckets at any instant, and does 
the work of overcoming P lt the resistance of the mill machinery, 
and also the friction. By placing P x pointing upward on the 
same side of C as P, and making h x nearly = h, R will = G 
nearly, just as when the wheel is running empty; and the 
foregoing numerical results will still hold good for practical 
purposes. 

168. Friction of Pivots. — In the case of a vertical shaft or 
axle, and sometimes in other cases, the extremity requires sup- 
port against a thrust along the axis of the axle or pivot. If 
the end of the pivot isjlat and also the surface 
against which it rubs, we may consider the 
pressure, and therefore the friction, as uniform 
over the surface. With a flat circular pivot, 
then, Fig. 180, the frictions on a small sector 
of the circle form a system of parallel forces 
whose resultant is equal to their sum, and is FlG * 18 °* 
applied a distance of \r from the centre. Hence the sum of 
the moments of all the frictions about the centre =fP%r, in 
which R is the axial pressure. Therefore a force P necessary 
to overcome the friction with uniform rotation must have a 
moment 

Pa =fBfr, 




180 MECHANICS OF ENGINEERING. 

and the work lost in friction per revolution is 

=/!»*. -Ir-fVKr (1) 

As the pivot and step become worn, the resultant frictions 
in the small sectors probably approach the centre ; for the 
greatest wear occurs first near the outer edge, since there the 
product^ is greatest (see § 166). Hence for \r we may more 
reasonably put Jr. 

Example. — A vertical flat-ended pivot presses its step with 
a force of 12 tons, is 6 inches in diameter, and makes 40 revolu- 
tions per minute. Required the H. P. absorbed by the friction. 
Supposing the pivot and step new, and f for good lubrication 
= 0.07, we- have, from eq. (1) {foot-lb. -second) , 

Work lost per revolution 

= .07 X 21,000 X 6.28 X * . i = 1758.4 ft.-lbs., 

and .'. work per second 

= 1758.4 XU = 1H2.2 ft.-lbs., 

which -7-550 gives 2.13 H. P. absorbed in friction. If ordi- 
nary axle-friction also occurs its effect must be added. 

If the flat-ended pivot is hollow, with radii r 1 and /•„ we may 
put ^(r l -f- r t ) instead of the \r of the preceding. 

It is obvious that the smaller the lever-arm given to the 
resultant friction in each sector of the rubbing surface the 
smaller the power lost in friction. Hence pivots should be- 
made as small as possible, consistently with strength. 

For a conical pivot and step, Fig. 181, the resultant friction 
in each sector of the conical bearing surface has 
a lever-arm = %r x about the axis A, and a value 
> than for a flat-ended pivot ; for, on account 
of the wedge-like action of the bodies, the 
pressure causing friction is greater. The sum of 
the moments of these resultant frictions about 
A is the same as if only two elements of the 
cone received pressure (each = N — \R -5- sin a). Hence the 




FKICTION. 



181 



moment of friction of the pivot, i.e., the moment of the force 
necessary to maintain uniform rotation, is 

Pa =f2JV^r 1 =-f-^— |n, 
J 3 l J sin a 3 " 

and work lost per revolution = ft 7 */- ?V 

r 3 ^ sin a * 

By making t\ small enough, these values may be made less 
than those for a flat-ended pivot of the same diameter — 2r. 

In Schiele's "anti-friction" pivots the outline is designed 
according to the following theory for securing uniform vertical 
wear. Let^> = the pressure per 
horizontal unit of area (i.e., 
= i? -T- horizontal projection of ^ 
the actual rubbing surface) ; 
this is assumed constant. Let 
the unit of area be small, for 
algebraic simplicity. The fric- FlG . i82. 

tion on the rubbing surface, whose horizontal projection = unity, 
is = fN =f (p -+- sin a) (see Fig. 182; the horizontal com- 
ponent of p is annulled by a corresponding one opposite). The 
work per revolution in producing wear on this area = fN2ny. 
But the vertical depth of wear per revolution is to be the same 
at all parts of the surface ; and this implies that the same 
volume of material is worn away under each horizontal unit of 

V 

area. Hence fN^ny, i.e.,/ *. 2?n/, is to be constant for all 

sin a 

values of y ; and since jj? and 2n are constant, we must have, 

as the law of the curve, 

y 




sin a 



i.e., the tangent BC = the same at all points. 



This curve is called the " tractrixP Schiele's pivots give a 
very uniform wear at high speeds. The smoothness of wear 
prevents leakage in the case of cocks and faucets. 

169. Normal Pressure of Belting. — When a perfectly flexible 
cord, or belt, is stretched over a smooth cylinder, both at rest, 



182 



MECHANICS OF ENGINEERING. 



the action between them is normal at every point. As to its 
j&- \\\\ s amount, p, per linear unit of arc, the fol- 
£*V*^=?r ** lowing will determine. Consider a semi- 
circle of the cord free, neglecting its weight. 
Fig. 183. The force holding it in equilib- 
rium are the tensions at the two ends (these 
are equal, manifestly, the cylinder being 
smooth ; for they are the only two forces 
""* having moments about C, and each has the 
same lever-arm), and the normal pressures, 
which are infinite in number, but have an intensity, p, per 
linear unit, which must be constant along the curve since S is 
the same at all points. The normal pressure on a single ele- 
ment, ds, of the cord is = pds, and its X component =. 
pds cos 6 = prdd cos 0. Hence 2X = gives 




cos Odd — 2S=0, i.e., rp\ sin 6 = 28; 

i* L- in 



rp\l — (— 1)] = 2S or p = 



S 



(1) 



170. Belt on Rough Cylinder. Impending Slipping. — If fric- 
tion is possible between the two bodies, the tension may vary 
along the arc of contact, so that p also varies, and consequently 




Fia. 184. 




the friction on an element ds being =Jpds =zf{S-r- r)ds, also 
varies- If slipping is impending, the law of variation of the 
tension S may be found, as follows : Fig. 184, in which the 



FRICTION. 183 

impending slipping is toward the left, shows the cord free. 
For any element, ds, of the cord, we have, putting 2 (moms, 
about O) = (Fig. 185), 

(S+ dS)r = Sr + dFr ; i.e., dF= dS, 

or (see above) dS =f(S -i- r)ds. 

But ds = rdd ; hence, after transforming, 

/^ = f (1) 

In (1) the two variables and S are separated ; (1) is there- 
fore ready for integration. 

fa = log, S n - log e S, = lo ge [|]. (2) 

Or, by inversion, S ef a — S n , (3) 

e, denoting the Kaperian base, = 2.71828 -fs a of course is in 
TT-measure. 

Since S n evidently increases very rapidly as a becomes 
larger, S remaining the same, we have the explanation of the 
well-known fact that a comparatively small tension, # , exerted 
by a man, is able to prevent the slipping of a rope around a 
pile-head, when the further end is under the great tension S n 
due to the stopping of a moving steamer. For example, with 
f = ^, we have (Weisbach) 

for a = J turn, or a — \n, S n = 1.69# 

= \ turn, or a — 7T, S n — 2.&58 

= 1 turn, or a — 2?r, S n = 8.12# 

= 2 turns, or a = 4?r, S n = 65.94# ; 

= 4 turns, or a = 8tt, S n = 4348.56£ . ' 

If slipping actually occurs, we must use a value of f for fric- 
tion of motion. 

Example. — A leather belt drives an iron pulley, covering 
one half the circumference. What is the limiting value of the 



184 



MECHANICS OF ENGINEEKING. 



ratio of S n (tension on driving-side) to S (tension on follow- 
ing side) if the belt is not to slip, taking the low value of 
f = 0.25 for leather on iron ? 

We have given fa = 0.25 X tt = .7854, which by eq. (2) is 
the Naperian log. of (S n : S ) when slipping. occurs. Hence the 
common log. of (S n : S ) = 0.7854 X 0.43429 = 0.34109 ; i.e., 
if 

(S n : S ) = 2.193, say 2.2, 
the belt will slip (for/= 0.25). 

(0.43429 is the modulus of the common system of loga- 
rithms, and = 1 : 2.30258. See example in §48.) 

At very high speeds the relation^? = xS Y -=- r (in § 169) is not 
strictly true, since the tensions at the two ends of an element 
ds are partly employed in furnishing the necessary deviating 
force to keep the element of the cord in its circular path, the 
remainder producing normal pressure. 

171. Transmission of Power by Belting or Wire Rope. — In the 

simple design in Fig. 186, it is required to find the motive 
weight G, necessary to overcome the given resistance JR at a 



DRIVING SIDE 




Fig. 186. 



uniform velocity = v x \ also the proper stationary tension 
weight G to prevent slipping of the belt on its pulleys, and 
the amount of power, Z, transmitted. 
In other words, 



Given : 



( i?, a, r, a 1 ,r 1 ;a = n for both pulleys, ) 
( «>,; and f for both pullevs ; 



and/ for both pulleys ; 

-p . , , ( L ; G, to furnish L ; G for no slip ; v the velocity 
* \ of G ; v' that of belt ; and the tensions in belt. 



FRICTION. 



185 



Neglecting axle-friction and the rigidity of the belting, the 
power transmitted is that required to overcome R through a 
distance = v x every second, i.e., 

Z = Rv x (1) 

Since (if the belts do not slip) 

a : r : : v' : v, and a x : r x : : v' : v x , 



we have 



, a x r a x 

v = —v„ and v = v.. 

r x " a r x x 



(2) 



Neglecting the mass of the belt, and assuming that each pul- 
ley revolves on a gravity-axis, we obtain the following, by con- 
sidering the free bodies in Fig. 187 : 




CA free) 



(B free) 
Fig.* 187. 



(B and" tcu.ck fisg) 



2 (moms.) = in A free gives Rr x = (S n — S )a x ; . (3) 
2 (moms.) = in B free gives Gr = (S n — S )a ; . (4) 



whence we readily find 



r a, 



Evidently R and G are inversely proportional to their velo- 
cities v x and v ; see (2). This ought to be true, since in Fig. 
186 G is the only working-force, R the only resistance, and 
the motions are uniform ; hence (from eq. (XYL), § 142) 

Gv - Rv x = 0. 

2X = 0, for B and truck free, gives 

G 9 = Sn + S„ (5) 

while, for impending slip, 

S n = S e f " (6) 



186 MECHANICS OF ENGINEERING. 

By elimination between (4), (5), and (6), we obtain 

^° ~~ a 'ef« — l~v''ef* — V ' ' ' Ki) 
L eJ n 

and S " = v" W^\ ( 8 ) 

Hence G and S n vary directly as the power transmitted and 
inversely as the velocity of the belt. For safety G should be 
made > the above value in (7) ; corresponding values of the 
two tensions may then be found from (5), and from the rela- 
tion (see § 150) 

(S n -S )v' = Z (6a) 

These new values of the tensions will be found to satisfy the 
condition of no slip, viz., 

(£„:£„)< ^(§170). 

For leather on iron, e fir = 2.2 (see example in § 170), as a 
low value. The belt should be made strong enough to with- 
stand S n safely. 

As the belt is more tightly stretched, and hence elongated, 
on the driving than on the following side, it " creeps" back- 
ward on the driving and forward on the driven pulley, so that 
the former moves slightly faster than the latter. The loss of 
work due to this cause does not exceed 2 percent with ordinary 
belting (Cotterill). 

In the foregoing it is evident that the sum of the tensions in 
the two sides = 6r , i.e., is the same, whether the power is 
being transmitted or not ; and this is found to be true, both in 
theory and by experiment, when a tension- weight is not used, 
viz., when an initial tension S is produced in the whole belt 
before transmitting the power, then after turning on the latter 
the sum of the two tensions (driving and following) always 
= 2/S, since one side elongates as much as the other contracts ; 
it being understood that the pulley-axles preserve a constant 
distance apart. 

172. Rolling Friction. — The few experiments which have 
been made to determine the resistance offered by a level road- 




FRICTION. 187 

way to the uniform motion of a roller or wheel rolling upon it 
corroborate approximately the following theory. The word 
friction is hardly appropriate in this connection (except when 
the roadway is perfectly elastic, as will be seen), but is sanctioned 
by usage. 

First, let the roadway or track be compressible, but inelastic, 
G the weight of the roller and its load, and P the horizontal 
force necessary to preserve a uniform motion 
(both of translation and rotation). The track 
(or roller itself) being compressed just in 
front, and not reacting symmetrically from 
behind, its resultant pressure against the 
roller is not at vertically under the centre, FlG " 188 ' 

but some small distance, OD == b, in front. (The successive 
crushing of small projecting particles has the same effect.) 
Since for this case of motion the forces have the same relations 
as if balanced (see § 124), we may put 2 moms, about D = 0, 

.:Pr=Gb; or, P = l -0 (1) 

Coulomb found for 

Rollers of lignum-vitse on an oak track, b = 0.0189 inches; 
Hollers of elm on an oak track, b = 0.0320 inches. 

Weisbach's experiments give, for cast-iron wheels 20 inches in 
diameter on cast-iron rails, 

b = 0.0183 inches ;, 
and Pittinger, for the same, b = 0.0193 inches. 

Pambour gives, for iron railroad wheels 39.4 inches in diameter, 

I = 0.0196 to 

0.0216 inches. 

According to the foregoing theory, JP, the " rolling friction" 
(see eq. (1)), is directly proportional to G, and inversely to the 
radius, if b is constant. The experiments of General Morin and 
others confirm this, while those of Dupuit, Poiree. and Sauvage 
indicate it to be proportional directly to G, and inversely to the 
square root of the radius. 



188 MECHANICS OF ENGINEERING. 

Although b is a distance to be expressed in linear units, and 
not an abstract number like the f and f for sliding and axle- 
friction, it is sometimes called a " coefficient of rolling fric- 
tion." In eq. (1), b and r should be expressed in the same 
unit. 

Of course if P is applied at the top of the roller its lever- 
arm about D is 2r instead of r, with a corresponding change 
in eq. (1). 

With ordinary railroad cars the resistance due to axle and 
rolling frictions combined is about 8 lbs. per ton of weight on 
a level track. For wagons on macadamized roads b = \ inch, 
but on soft ground from 2 to 3 inches. 

Secondly, when the roadway is perfectly elastic. This is 
chiefly of theoretic interest, since at first sight no force would 
be considered necessary to maintain a uniform rolling motion. 
But, as the material of the roadway is compressed under the 
roller its surface is first elongated and then recovers its former 
state ; hence some rubbing and consequent sliding friction must 




Fig. 189. 



occur. Fig. 189 gives an exaggerated view of the circum- 
stances, P being the horizontal force applied at the centre 
necessary to maintain a uniform motion. The roadway (rub- 
ber for instance) is heaped up both in front and behind the 
roller, being the point of greatest pressure and elongation 
of the surface. The forces acting are G, P, the normal 
pressures, and the frictions due to them, and must form a 
balanced system. Hence, since G and P, and also the normal 
pressures, pass through C, the resultant of the frictions must 
also pass through C\ therefore the frictions, or tangential 
actions, on the roller must be some forward and some backward 



FRICTION. 



189' 



(and not all in one direction, as seems to be asserted on p. 260 
of Cotterill's Applied Mechanics, where Professor Beynolds' 
explanation is cited). The resultant action of the roadway 
upon the roller acts, then, through some point J), a distance 
OD = b ahead of 0, and in the direction DC, and we have as. 
before, with D as a centre of moments, 



Pr = Gb, 



or 



P = -G. 

r 



If rolling friction is encountered above as 
well as below the rollers, Fig. 190, the 
student may easily prove, by considering 
three separate free bodies, that for uniform 
motion 



= h 4^e, 



2r 




where b and b x are the respective " coefficients of rolling fric- 
tion" for the upper and lower contacts. 

Example 1. — If it is found that a train of cars will move- 
uniformly down an incline of 1 in 200, gravity being the only 
working force, and friction (both rolling and axle) the only- 
resistance, required the coefficient, f\ of axle-friction, the' 
diameter of all the wheels being 2r = 30 inches, that of the 
journals 2a = 3 inches, taking b = 0.02 inch for the rolling- 
friction. Let us use equation (XYI.) (§142), noting that while 
the train moves a distance s measured on the incline, its weight 

G does the work G -^-x s, the rolling friction — G (at the axles) 

has been overcome through the distance s, and the axle-friction 

(total) through the (relative) distance - s in the journal boxes ;.. 

whence, the change in kinetic energy being zero, 

1 „ b 



200 



Gs- Gs--fGs=0. 



Gs cancels out, the ratios b : r and a : r are 



Tsinr an d tV 



respectively (being ratios or abstract numbers they have the 



190 MECHANICS OF ENGINEERING. 

same numerical values, whether the inch or foot is used), and 
solving, we have 

/ = 0.05 - 0.0133 = 0.036. 
Example 2. — How many pounds of tractive effort per ton 
of load would the train in Example 1 require for uniform mo- 
tion on a level track % Arts. 10 lbs. 

173. Railroad Brakes. — During the uniform motion of a 
railroad car the tangential action between the track and each 
wheel is small. Thus, in Example 1, just cited, if ten cars of 
eight wheels each make up the train, each car weighing 20 tons, 
the backward tangential action of the rails upon each wheel is 
only 25 lbs. When the brakes are applied to stop the train 
this action is much increased, and is the only agency by which 
the rails can retard the train, directly or indirectly : directly, 
when the pressure of the brakes is so great as to prevent the 
wheels from turning, thereby causing them to "skid" (i.e., 
slide) on the rails ; indirectly, when the brake-pressure is of 
such a value as still to permit perfect rolling of the wheel, in 
which case the rubbing (and heating) occurs between the brake 
and wheel, and the tangential action of the rail has a value 
equal to or less than the friction of rest. In the first case, 
then (skidding), the retarding influence of the rails is the fric- 
tion of motion between rail and wheel ; in the second, a force 
which may be made as great as the friction of rest between rail 
and wheel. Hence, aside from the fact that skidding produces 
objectionable flat places on the wheel-tread, the brakes are 
more effective if so applied that skidding is impending, but 
not actually produced ; for the friction of rest is usually greater 
than that of actual slipping (§ 160). This has been proved 
experimentally in England. The retarding effect of axle and 
rolling friction has been neglected in the above theory. 

Example 1. — A twenty-ton car with an initial velocity of 80 
feet per second (nearly a mile a minute) is to be stopped on a 
level within 1000 feet ; required the necessary friction on each 
of the eight wheels. 

Supposing the wheels not to skid, the friction will occur 



FRICTION. 191 

between the brakes and wheels, and is overcome through the 
(relative) distance 1000 feet. Eq. (XYL), § 142, gives (foot- 
lb.-second system) 

O-SFX 1000 = - j ^Jc 80 )'* 

from which F ( = friction at circumference of each wheel) 

= 496 lbs. 

Example 2. — Suppose skidding to be impending in the fore- 
going, and the coefficient of friction of rest (i.e., impending 
slipping) between rail and wheel to be/* =0.20. In what 
distance will the car be stopped ? Ans. 496 ft. 

Example 3. — Suppose the car in Example 1 to be on an up- 
grade of 60 feet to the mile. (In applying eq. (XYI.) here, 
the weight 20 tons will enter as a resistance.) Ans. 439 lbs. 

Example 4. — In Example 3, consider all four resistances, 
viz., gravity, rolling friction, and brake and axle frictions, the 
distance being 1000 ft., and i^the unknown quantity. 

Ans. 414 lbs. 

174. Estimation of Engine and Machinery Friction. — Accord- 
ing to Professor Cotterill, a convenient way of estimating the 
work lost in friction in a steam-engine and machinery driven 
by it is the following : 

Let p m = mean effective steam-pressure per unit of area of 
piston, and conceive this composed of three por- 
tions, viz., 
p — the necessary pressure to drive the engine alone un- 
loaded, at the proper speed ; 
j}' m = pressure necessary to overcome the resistance caused 

by the useful work of the machines ; 
ep ' m == pressure necessary to overcome the friction of the 
machinery, and that of the engine over and above 
its friction when unloaded. This is about 15$ of 
p' m (i.e., e = 0.15), except in large engines, and 
then rather less. 
That is, by formula, F being the piston- area and I the length 
of stroke, the work per stroke is thus distributed : 



192 MECHANICS OF ENGINEERING. 

p is " from 1 to 1J-, or in marine engines 2 lbs. or more per 
square inch." 

175. Anomalies in Friction. — Experiment has shown consid- 
erable deviation under certain circumstances from the laws of 
friction, as stated in § 157 for sliding friction. At pressures 
below f lb. per sq. inch the coefficient f increases when the- 
pressure decreases, while above 500 lbs. (Kennie, with iron and 
steel) it increases with the pressure. "With high velocities, how- 
ever, above 10 ft. per second, f is much smaller as the velocity 
increases (Bochet, 1858). 

As for axle-friction, experiments instituted by the Society of 
Mechanical Engineers in England (see the London Engineer 
for March 7 and 21, 1884) gave values for f less than -^ 
when a "bath" of the lubricant was employed. These values 
diminished with increase of pressure, and increased with the 
velocity (see below, Hirn's statement). 

Professor Cotterill says, " It cannot be doubted that for 
values of pv (see § 166) > 5000 the coefficient of friction of 
well-lubricated bearings of good construction diminishes with 
the pressure, and may be much less than the value at low speeds 
as determined by Morin" (p. 259 of his Applied Mechanics). 

Professor Thurston's experiments confirmed those of Hirn as 
to the following relation : "The friction of lubricated surfaces 
is nearly proportional to the square root of the area and pres- 
sure." Hirn also maintained that, " in ordinary machinery, 
friction varies as the square root of the velocity." 

176. Rigidity of Ropes. — If a rope or wire cable passes over 
a pulley or sheave, a force P is required on one side greater 
than the resistance Q on the other for uniform motion, aside 
from axle-friction. Since in a given time both P and Q 
describe the same space s, if P is > Q, ^ ien Psh > Qs, i.e., 
the work done by P is > than that expended upon Q. This 
is because some of the work Ps has been expended in bending 
the stiff rope or cable, and in overcoming friction between the 
strands, both where the rope passes upon and where it leaves- 



FRICTION. 



193 



191, the material being 



the pulley. With hemp ropes, Fig, 

nearly inelastic, the energy spent in bending it on at D is 

nearly all lost, and energy must also be spent in straightening 




Fig. 191. 



it at E\ but with a wire rope or cable some of this energy is 
restored by the elasticity of the material. The energy spent 
in friction or rubbing of strands, however, is lost in both cases. 
The figure shows geometrically why P must be > Q for a 
uniform motion, for the lever-arm, a, of P is evidently < h 
that of Q. If axle-friction is also considered, we must have 

Pa=Qb+f(P+Q)r, 

r being the radius of the journal. 

Experiments with cordage have been made by Prony, Cou- 
lomb, Eytelwein, and Weisbach, with considerable variation in 
the results and formulse proposed. (See Coxe's translation of 
vol. i., Weisbach's Mechanics.) 

With pulleys of large diameter the effect of rigidity is very 
slight. For instance, Weisbach gives an example of a pulley 
five feet in diameter, with which, Q being = 1200 lbs., P 
= 1219. A wire rope f in. in diameter was used. Of this 
difference, 19 lbs., only 5 lbs. was due to rigidity, the remainder, 
14 lbs., being caused by axle-friction. When a hemp-rope 1.6 
inches in diameter was substituted for the wire one, P— Q=27 
lbs., of which 12 lbs. was due to the rigidity. Hence in one 
case the loss of work was less than -J- of 1%, in the other about 
1$, caused by the rigidity. For very small sheaves and thick 
ropes the loss is probably much greater. 
13 



194 MECHANICS OF ENGINEERING. 

177. Miscellaneous Examples. — Example 1. The end of a 

shaft 12 inches in diameter and making 50 revolutions per min- 
ute exerts against its bearing an axial pressure of 10 tons and 
a lateral pressure of 40 tons. With f —f — 0.05, required 
the H. P. lost in friction. Arts. 22.2 H. P. 

Example 2. — A leather belt passes over a vertical pulley, 
covering half its circumference. One end is held by a spring 
balance, which reads 10 lbs. while the other end sustains a 
weight of 20 lbs., the pulley making 100 revolutions per min- 
ute. Required the coefficient of friction, and the H. P. spent 
in overcoming the friction. Also suppose the pulley turned 
in the other direction, the weight remaining the same. The 
diameter of the pulley is 18 inches. . (f = 0.22 ; 

m ' \ 0.142 and 284 H. P. 

Example 3. — A grindstone with a radius of gyration = 12 
inches has been revolving at 120 revolutions per minute, and 
at a given instant is left to the influence of gravity and axle 
friction. The axles are 1-J inches in diameter, and the wheel 
makes 160 revolutions in coming to rest. Required the coeffi- 
cient of axle-friction. Ans. f '= 0.389. 

Example 4. — A board A, weight 2 lbs., rests horizontally on 
another B\ coefficient of friction of rest between them being 
f — 0.30. B is now moved horizontally with a uniformly 
accelerated motion, the acceleration being = 15 feet per " square 
second ;" will A keep company with it, or not ? Ans. " No." 



PART III. 

STKENGTH OF MATERIALS. 

[Or Mechanics of Materials]. 



CHAPTER I. 

ELEMENTARY STRESSES AND STRAINS. 

178. Deformation of Solid Bodies. — In the preceding por- 
tions of this work, what was called technically a " rigid 
body," was supposed incapable of changing its form, i.e., 
the positions of its particles relatively to each other, under 
the action of any forces to be brought upon it. This sup- 
position was made because the change of form which must 
actually occur does not appreciably alter the distances, 
angles, etc., measured in any one body, among most of 
the pieces of a properly designed structure or machine. 
To show how the individual pieces of such constructions 
should be designed to avoid undesirable deformation or 
injury is the object of this division of Mechanics of En- 
gineering, viz., the Strength of Materials. 



Fig. 192. § 178. 

As perhaps tne simplest instance of the deformation or 
distortion of a solid, let us consider the case of a prismatic 
rod in a state of tension, Fig. 192 (link of a surveyor's 



196 MECHANICS OF ENGINEERING. 

chain, e.g.). The pull at each end is P, and the body is 
said to be tinder a tension of P (lbs., tons, or other unit), 
not 2P. Let ABGD be the end view of an elementary 
parallelopiped, originally of square section and with faces 
at 45° with the axis of the prism. It is now deformed, the 
four faces perpendicular to the paper being longer than 
before, while the angles BAD and BCD, originally right 
angles, are now smaller by a certain amount o, ABC and 
ADC larger by an equal amount d. The element is said 
to be in a state of strain, viz.: the elongation of its edges 
(parallel to paper) is called a tensile strain, while the alter- 
ation in the angles between its faces is called a shearing 
strain, or angular distortion (sometimes also called a slid- 
ing, or tangential, strain, since BG has been made to slide, 
relatively to AD, and thereby caused the change of angle). 
[This use of the word strain, to signify change of form and 
not the force producing it, is of recent adoption among 
many, though not all, technical writers.] 

179. Strains. Two Kinds Only. — Just as a curved line may 
be considered to be made up of small straight-line ele- 
ments, so the substance of any solid body may be consid- 
ered to be made up of small contiguous parallelopipeds, 
whose angles are each 90° "before the body is subjected to 
the action of forces, but which are not necessarily cubes. 
A line of such elements forming an elementary prism is 
sometimes called a fibre, but this does not necessarily imply 
a fibrous nature in the material in question. The system 
of imaginary cutting surfaces by which the body is thus 
subdivided need not consist entirely of planes ; in the sub- 
ject of Torsion, for instance, the parallelopipedical ele- 
ments considered lie in concentric cylindrical shells, cut 
both by transverse and radial planes. 

Since these elements are taken so small that the only 
possible changes of form in any one of them, as induced 
by a system of external forces acting on the body, are 



ELEMENTARY STRESSES, ETC. 197 

elongations or contractions of its edges, and alteration of 
its angles, there are but two kinds of strain, elongation 
(contraction, if negative) and shearing, 

180. Distributed Forces or Stresses. — In the matter preced- 
ing this chapter it has sufficed for practical purposes to 
consider a force as applied at a point of a body, but in 
reality it must be distributed over a definite area ; for 
otherwise the material would be subjected to an infinite 
force per unit of area. (Forces like gravity, magnetic at- 
traction, etc., we have already treated as distributed over 
the mass of a body, but reference is now had particularly 
to the pressure of one body against another, or the action 
of one portion' of the body on the remainder.) For in- 
stance, sufficient surface must be provided between the 
end of a loaded beam and the pier on which it rests to 
avoid injury to either. Again, too small a wire must not 
be used to sustain a given load, or the tension per unit 
of area of its cross section becomes sufficient to rupture 
it. 

Stress is distributed force, and its intensity at any point 
of the area is 

-& ■ ■ ■ « 

where dF is a small area containing the point and dP the 
force coming upon that area. If equal dP's (all parallel) 
act on equal dF'soi a plane surface, the stress is said to 
be of uniform intensity, which is then 

P= P F .... (2) 

where P= total force and F the total area over which it 
acts. The steam pressure on a piston is an example of 
stress of uniform intensity. 



198 MECHANICS OF ENGINEERING. 

For example, if a force P= 28800 lbs, is uniformly dis- 
tributed over a plane area of F= 72 sq. inches, or y 2 of a 
sq. foot, the intensity of the stress is 

28800 , nnl , . , 

=400 lbs. per sq. men, 



72 

(or p = 28800-5- j£ =57600 lbs. per sq. foot, or £>=14.400+ 
^=28.8 tons per sq. ft., etc.). 



181. Stresses on an Element ; of Two Kinds Only. — When a 
solid body of any material is in equilibrium under a sys- 
tem of forces which do not rupture it, not only is its shape 
altered (i.e. its elements are strained), and stresses pro- 
duced on those planes on which the forces act, but other 
stresses also are induced on some or all internal surfaces 
which separate element from element, (over and above the 
forces with which the elements may have acted on each 
other before the application of the external stresses or 
" applied forces "). So long as the whole solid is the "free 
body " under consideration, these internal stresses, being 
the forces with which the portion on one side of an imag- 
inary cutting plane acts on the portion on the other side, 
do not appear in any equation of equilibrium (for if intro- 
duced they would cancel out); but if we consider free a 
portion only, some or all of whose bounding surfaces are 
cutting planes of the original body, the stresses existing 
on these planes are brought into the equations of equilib- 
rium. 

Similarly, if a single element of the body is treated by 
itself, the stresses on all six of its faces, together with its 
weight, form a balanced system of forces, the body being 
supposed at rest. 





Fig. 193. 



ELEMEXTARY STRESSES, ETC. 199 

As an example of internal stress, consider again the case 
<of a rod in tension ; Fig. 193 shows the whole rod (or eye- 
bar) free, the forces P being the pressures of the pins in 
the eyes, and causing external stress (compression here) 
on the surfaces of contact. Conceive a right section made 
through BS, far enough from the eye, (7, that we may con- 
sider the internal stress to be uniform in this section, and 
consider the portion BSG as a free body, in Fig. 194. The 
stresses on BS, now one of the bounding surfaces of the 
free body, must be parallel to P, i.e., normal to BS ; 
(otherwise they would have components perpendicular to 
P, which is precluded by the necessity of 17 being = 0, 
and the supposition of uniformity.) Let F = the sec- 



FlG. 194. 




Fig. 195. 

tional area BS, and p = the stress per unit of area ; then 

IX= gives P= Fp, i.e., p=? . . (2) 

F 

The state of internal stress, then, is such that on planes 
perpendicular to the axis of the bar the stress is tensile and 
normal (to those planes). Since if a section were made 
oblique to the axis of the bar, the stress would still be 
parallel to the axis for reasons as above, it is evident that 
on an oblique section, the stress has components both nor- 
mal and tangential to the section, the normal component 
being a tension. 



200 



MECHANICS OF ENGINEERING. 



The presence of the tangential or shearing stress in ob- 
lique sections is rendered evident by considering that if an 
oblique dove-tail joint were cut in the rod, Fig. 195, the 
shearing stress on its surfaces may be sufficient to over- 
come friction and cause sliding along the oblique plane. 

If a short prismatic block is under the compressive ac- 
tion of two forces, each = P and applied centrally in one 
base, we may show that the state of internal stress is the 
same as that of the rod under tension, except that the nor- 
mal stresses are of contrary sign, i.e., compressive instead 
of tensile, and that the shearing stresses (or tendency to 
slide) on oblique planes are opposite in direction to those 
in the rod. 

Since the resultant stress on a given internal plane of a 
body is fully represented by its normal and tangential 
components, we are therefore justified in considering but 
two kinds of internal stress, normal or direct, and tangen- 
tial or shearing. 

182. Stress on Oblique Section of Rod in Tension. — Consider 
b free a small cubic element whose 

edge =a in length; it has two 
faces parallel to the paper, being 
taken near the middle of the rod 
in Fig. 192. Let the angle which 
the face AB, Fig. 196, makes with 
the axis of the rod be = a. This 
angle, for our present purpose, is 
considered to remain the same 
while the two forces P are acting, 
as before their action. The re- 
sultant stress on the face AB hav- 
ing an intensity p=P~F, (see eq. 
2) per unit of transverse section 
of rod, is = p (a sin a) a. Hence 
FlG - m - its component normal to AB is 

pa 2 sin 2 a ; and the tangential or shearing component along 



















—/^ w \ 


r 


A^ 




y 










c^ 


















\n^^ 








ELEMENTARY STRESSES, ETC. 201 

AB=pa 2 sin a cos a. Dividing by the area, a 2 , we have 
the following : 

For a rod in simple tension we have, on a plane making 
an angle, a, with the axis : 

a Normal Stress =p sin 2 a per unit of area . . (1) 
and a Shearing Stress =p sin a cos a per unit of area . (2) 

" Unit of area " here refers to the oblique plane in ques- 
tion, while p denotes the normal stress per unit of area of 
a transverse section, i.e., when a=90°, Fig. 194. 

The stresses on CD are the same in value as on AB, 
while for BG and A I) we substitute 90° — a for a. Fig. 
197 shows these normal and shearing stresses, and also, 
much exaggerated, the strains or change of form of the 
element (see Fig. 192). 

182a. Relation between Stress and Strain. — Experiment 
shows that so long as the stresses are of such moderate 
value that the piece recovers its original form completely 
when the external forces which induce the stresses are re- 
moved, the following is true and is known as Hooke's Law 
(stress proportional to strain). As the forces P in Fig. 
193 (rod in tension) are gradually increased, the elonga- 
tion, or additional length, of RK increases in the same 
ratio as the normal stress, p, on the sections MS and KN, 
per unit of area [§ 191]. 

As for the distorting effect of shearing stresses, consider 
in Fig. 197 that since 

p sin a cos a = p cos (90° — a) sin (90° — a) 

the shearing stress per unit of area is of equal value on all 
four of the faces (perpendicular to paper) in the elementary 
block, and is evidently accountable for the shearing strain, 
i.e., for the angular distortion, or difference, d, between 
90° and the present value of each of the four angles. Ac- 
cording to Hooke's Law then, as P increases within tha 
limit mentioned above, o varies proportionally to 

p sin a cos «, i.e. to the stress. 



202 MECHANICS OF ENGINEERING. 

182b. Example. — Supposing the rod in question were of 
a kind of wood in which a shearing stress of 200 lbs. per 
sq. inch along the grain, or a normal stress of 400 lbs. per 
sq. inch, perpendicular to a fibre-plane will produce rup- 
ture, required the value of a the angle which the grain 
must make with the axis that, as P increases, the danger 
of rupture from each source may be the same. This re- 
quires that 200:400::^ sin a cos a:p sin 2 a, i.e. tan. a must 
= 2.000.-.«=63^°. If the cross section of the rod is 2 sq. 
inches, the force P at each end necessary to produce rup- 
ture of either kind, when «=63^°, is found by putting 
_psin a cos #= 200. \p =500.0 lbs. per sq. inch. Whence, since 
p=P- : rF ) P=1000 lbs. (Units, inch and pound.) 

183. Elasticity is the name given to the property which 
most materials have, to a certain extent, of regaining their 
original form when the external forces are removed. If 
the state of stress exceeds a certain stage, called the Elastic 
Limit, the recovery of original form on the part of the ele- 
ments is only partial, the permanent deformation being 
called the Set. 

Although theoretically the elastic limit is a perfectly defi- 
nite stage of stress, experimentally it is somewhat indefi- 
nite, and is generally considered to be reached when the 
permanent set becomes well marked as the stresses are in- 
creased and the test piece is given ample time for recovery 
in the intervals of rest. 

The Safe Limit of stress, taken well within the elastic 
limit, determines the working strength or safe load of the 
piece under consideration. E.g., the tables of safe loads 
of the rolled wrought iron beams, for floors, of the New 
Jersey Steel and Iron Co., at Trenton, are computed on 
the theory that the greatest normal stress (tension or com- 
pression) occurring on any internal plane shall not exceed 
12,000 lbs. per sq. inch ; nor the greatest shearing stress 
4,000 lbs. per sq. inch. 



ELEMENTARY STRESSES, ETC. 203 

The Ultimate Limit is reached when rupture occurs. 

184. The Modulus of Elasticity (sometimes called co-efficient 
of elasticity) is the number obtained by dividing the stress 
per unit of area by the corresponding relative strain. 

Thus, a rod of wrought iron y 2 sq. inch sectional area 
being subjected to a tension of 2^ tons =5,000 lbs., it is 
found that a length which was six feet before tension is 
= 6.002 ft. during tension. The relative longitudinal strain 
or elongation is then= (0.002)-^ 6=1 : 3,000 and the corres- 
ponding stress (being the normal stress on a transverse 
plane) has an intensity of 

p t =P^- ^='5,000-^-^=10,000 lbs., per sq. inch. 

Hence by definition the modulus of elasticity is (for ten- 
sion) 

^ t =^ t ye=10,000-T- JL =30,000,000 lbs. per sq. inch, (the 

sub-script "t " refers to tension). 

It will be noticed that since e is an abstract number, E t 
is of the same quality as p ti i.e., lbs. per sq. inch, or one di- 
mension of force divided by two dimensions of length. 
(In the subject of strength of materials the inch is the 
most convenient English linear unit, when the pound is 
the unit of force ; sometimes the foot and ton are used to- 
gether.) 

The foregoing would be called the modulus of elasticity 
of wrought iron in tension in the direction of the fibre, as 
given by the experiment quoted. But by Hooke's Law_p 
and e vary together, for a given direction in a given ma- 
terial, hence within the elastic limit E is constant for a given 
direction in a given material. Experiment confirms this 
approximately. 

Similarly, the modulus of elasticity for compression E c 



204 MECHANICS OF ENGINEERING. 

in a given direction in a given material may be determined 
by experiments on short blocks, or on rods confined lat- 
erally to prevent flexure. 

As to the modulus of elasticity for shearing, E^ we 
divide the shearing stress per unit of area in the given 
direction by d (in tl measure) the corresponding angular 
strain or distortion ; e.g., for an angular distortion of 1° or 
#=.0174, and a shearing stress of 1,566 lbs. per sq. inch, 
we have ^=^=9,000,000 lbs. per sq. inch. 

Unless otherwise specified, by modulus of elasticity will 
be meant a value derived from experiments conducted 
within the elastic limit, and this, whether for normal stress 
or for shearing, is approximately constant for a given di- 
rection in a given substance.* 

185. Isotropes. — This name is given to materials which 
are homogenous as regards their elastic properties. In 
such a material the moduli of elasticity are individually 
the same for all directions. E.g., a rod of rubber cut out 
of a large mass will exhibit the same elastic behavior when 
subjected to tension, whatever its original position in the 
mass. Fibrous materials like wood and wrought iron are 
not isotropic ; the direction of grain in the former must 
always be considered. The " piling " and welding of nu- 
merous small pieces of iron prevent the resultant forging 
from being isotropic. 

186. Resilience refers to the potential energy stored in a 
body held under external forces in a state of stress which 
does not pass the elastic limit. On its release from con- 
straint, by virtue of its elasticity it can perform a certain 
amount of work called the resilience, depending in amount 
upon the circumstances of each case and the nature of the 
material. See § 148. 

187. General Properties of Materials. — In view of some defi- 
nitions already made we may say that a material is ductile 

* The moduli, or " co-efficients," of elasticity as used by physicists are well explained 
in Stewart and Gee's Practical Physics, Vol. I., pp. 164, eic. Their "co-efficient of 
rigidity" is our E s . 



ELEMENTARY STRESSES, ETC. 205 

when the ultimate limit is far removed from the elastic 
limit ; that it is brittle like glass and cast iron, when those 
limits are near together. A small modulus of elasticity 
means that a comparatively small force is necessary to 
produce a given change of form, and vice versa, but implies 
little or nothing concerning the stress or strain at the 
elastic limit ; thus Weisbach gives E c , lbs. per sq. inch for 
wrought iron =28,000,000= double the E c for cast iron 
while the compressive stresses at the elastic limit are the 
same for both materials (nearly). 

188. General Problem of Internal Stress. — This, as treated 
in the mathematical Theory of Elasticity, developed by 
Lame, Clapeyron and Poisson, may be stated as follows : 

Given the original form of a body when free from stress, 
and certain co -efficients depending on its elastic proper- 
ties ; required the neiv position, the altered shape, and the in- 
tensity of the stress on each of the six faces, of every element 
of the body, when a given balanced system of forces is applied 
to the body. 

Solutions, by this theory, of certain problems of the na- 
ture just given involve elaborate, intricate, and bulky 
analysis ; but for practical purposes Navier's theories 
(1838) and others of more recent date, are sufficiently exact, 
when their moduli are properly determined by experiments 
covering a wide range of cases and materials. These will 
be given in the present work, and are comparatively sim- 
ple. In some cases graphic will be preferred to analytic 
methods as more simple and direct, and indeed for some 
problems they are the only methods yet discovered for ob- 
taining solutions. Again, experiment is relied on almost 
exclusively in dealing with bodies of certain forms under 
peculiar systems of forces, empirical formulae being based 
on the experiments made ; e.g., the collapsing of boiler 
tubes, and in some degree the flexure of long columns. 



206 MECHANICS OF ENGINEERING. 

189. Classification of Cases. — Although in almost any case 
whatever of the deformation of a solid body by a balanced 
system of forces acting on it, normal and shearing stresses 
are both developed in every element which is affected at 
all (according to the plane section considered,) still, cases 
where the body is prismatic, and the external system con- 
sists of two equal and opposite forces, one at each end of 
the piece and directed away from each other, are commonly 
called cases of Tension; (Fig. 192); if the piece is a short 
prism with the same two terminal forces directed toward 
each other, the case is said to be one of Compression ; a case 
similar to the last, but where the prism is quite long 
(" long column "), is a case of Flexure or bending, as are also 
most cases where the " applied forces " (i.e., the external 
forces), are not directed along the axis of the piece. Rivet- 
ed joints and " pin-connections " present cases of Shearing; 
a twisted shaft one of Torsion. c When the gravity forces 
due to the weights of the elements are also considered, a 
combination of two or more of the foregoing general cases 
may occur. 

In each case, as treated, the principal objects aimed at 
are, so to design the piece or its loading that the greatest 
stress, in whatever element it may occur, shall not exceed 
a safe value ; and sometimes, furthermore, to prevent too 
great deformation on the part of the piece. The first ob- 
ject is to provide sufficient strength ; the second sufficient 
stiffness. 

190. Temperature Stresses. — If a piece is under such con- 
straint that it is not free to change its form with changes 
of temperature, external forces are induced, the stresses 
produced by which are called temperature stresses. 



TENSION. 



201 



TENSION. 

191. Hooke's Law by Experiment. — As atypical experiment 
in the tension of a long rod of ductile metal such as 
wrought iron and the mild steels, the following table is quot- 
ed from Prof. Cotterill's " Applied Mechanics." The experi- 
ment is old, made by Hodgkinson for an English Railway 
Commission, but well adapted to the purpose. From the 
great length of the rod, which was of wrought iron and 
0.517 in. in diameter, the portion whose elongation was 
observed being 49 ft. 2 in. long, the small increase in length 
below the elastic limit was readily measured. The succes- 
sive loads were of such a value that the tensile stress 
p=P^-F, or normal stress per sq. in. in the transverse 
section, was made to increase by equal increments of 2657.5 
lbs. per sq. in., its initial value. After each application of 
load the elongation was measured, and after the removal 
of the load, the permanent set, if any. 

Table of elongations of a wrought iron rod, of a lengtb— 49 it. 2 in. 



p 


; 


AX 


e=X+l 


X' 


Load, (lbs. per 
square inch.) 


Elongation, 
(inches.) 


Increment 

of 
Elongation. 


e, the relative 
elongation, (ab- 
stract number.) 


Permanent 

Set, 
(inches.) 


1X2667.5 


.0485 


.0485 


0.000082 




2X " 


.1095 


.061 


.000186 




3X " 


.1675 


.058 


.000283 


0.0015 


4X " 


.224 


.0565 


.000379 


.002 


5X " 


.2805 


.0565 


.000475 


.0027 


6X ll 


.337 


.0565 


.000570 


.003 


rx " 


.393 


.056 




.004 


8X " 


.452 


.059 


.000766 


.0075 


9X " 


.5155 


.0635 




.0195 


10X " 


.598 


.0825 




.049 


11X " 


.760 


.162 




.1545 


12X " 


1.310 


.550 




.667 


etc. 











208 



MECHANICS OE ENGINEERING. 



Keferring now to Fig. 198, the notation is evident. P 
is the total load in any experiment, F the cross section of 
the rod ; hence the normal stress on the transverse section 
is p=P-h-F. When the loads are increased by equal in- 
crements, the corresponding increments of the elongation 
k should also be equal if Hooke's law is true. It will be 
noticed in the table that this is very nearly true up to the 
8th loading, i.e., that Jk 9 the difference between two con- 
secutive values of k, is nearly constant. In other words the 
proposition holds good : 

p.p.-.-.ka, 



if P and P Y are any two loads below the 8th, and k and k x 
the corresponding elongations. 

The permanent set is just perceptible at the 3d load, and 
increases rapidly after the 8th, as also the increment of 
elongation. Hence at the 8th load, which produces a ten- 
sile stress on the cross section of p= 8x2667.5= 21340.0 
lbs. per sq. inch, the elastic limit is reached. 

As to the state of stress of the individual elements, if 

we conceive such sub -division 
of the rod that four edges of 
each element are parallel to the 
axis of the rod, we find that it 
is in equilibrium between two 
normal stresses on its end faces 
xaX (Fig. 199) of a value =pdF— 
(P^-F)dF where dF is the hor- 
izontal section of the element. 
If dx was the original length, 
and dk the elongation produced by pdF, we shall have, 
since all the dx's of the length are equally elongated at the 
same time, 

dk = k 

dx I 



dx 



a 



i 

Fig. 198. 




TENSION. 209 

where 1= total (original) length. But dl~-dx is the rela- 
tive elongation e, and by definition (§ 184) the Modulus of 
Elasticity for Tension, E l ,=p- : rs 

•••*=^rr OT *=j? • • • • w 

• i dx 

Eq. (1) enables us to solve problems involving the elonga- 
tion of a prism under tension, so long as the elastic limit 
is not surpassed. 

The values of E t computed from experiments like those 
just cited should be the same for any load under the elas- 
tic limit, if Hooke's law were accurately obeyed, but in 
reality they differ somewhat, especially if the material 
lacks homogeneity. In the present instance (see Table) 
we have from the 

2d Exper. E=p+-s= 28,680,000 lbs. per sq. in. 
5th " E= " =28,009,000 

8th " E t = " =27,848,000 

If similar computations were made beyond the elastic 
limit, i.e., beyond the 8th Exper., the result would be much 
smaller, showing the material to be yielding much more 
readily. 

192. Strain Diagrams. — If we plot the stresses per sq. inch 
(p) as ordinates of a curve, and the corresponding relative 
elongations (e) as abscissas, we obtain a useful graphic re- 
presentation of the results of experiment. 

Thus, the table of experiments just cited being utilized 
in this way, we obtain on paper a series of points through 
which a smooth curve may be drawn, viz. : OBC Fig. 200, 
for wrought iron. Any convenient scales may be used for 
p and e ; and experiments having been made on other 
metals in tension and the results plotted to the same scales 



210 



MECHANICS OF ENGINEERING. 



as before for p and e, we have the means of comparing their 
tensile properties. Fig. 200 shows two other curves, rep- 
resenting (roughly) the average behavior of steel and cast 
iron. At the respective elastic limits B f B', and B", it will 
be noticed that the curve for wrought iron makes a sudden 
turn from the vertical, while those of the others curve away 
more gradually ; that the curve for steel lies nearer the 
vertical axis than the others, which indicates a higher 
value for E t ; and that the ordinates BA', B'A', and B'A " 
(respectively 21,000, 9,000, and 30,000 lbs, per sq. inch) in- 




Fig. 200. 

dicate the tensile stress at the elastic limit. These latter 
quantities will be called the moduli of tenacity at elastic 
limit for the respective materials. [On a true scale the 
point G would be much further to the right than here 
shown. Only one half of the curve for steel is given, for 
want of space.] 

Within the elastic limit the curves are nearly straight 
(proving Hooke's law) and if «, «', and a" are the angles 
made by these straight portions with the axis of X (i.e., 
of e), we shall have 



(E t for w. iron) : (E t c. iron) : (E t steel) : : tan a : tan a' : tan a" 



TENSION. 211 

as a graphic relation between their moduli of elasticity 

(since E =%*-). 

s 

Beyond the elastic limit the wrought iron rod shows large 
increments of elongation for small increments of stress, 
i.e., the curve becomes nearly parallel to the horizontal 
axis, until rupture occurs at a stress of 53,000 lbs. per sq. 
inch of original sectional area (at rupture this area is some- 
what reduced, especially in the immediate neighborhood 
of the section of rupture ; see next article) and after a rel- 
ative elongation s= about 0.30, or 30%. (The preceding 
table shows only a portion of the results.) The curve 
for steel shows a much higher breaking stress (100,000 
lbs. per sq. in.) than the wrought iron, but the total 
elongation is smaller, e= about 10%. This is an average 
curve ; tool steels give an elongation at rupture of about 
4 to 5%, while soft steels resemble wrought iron in their 
ductility, giving an extreme elongation of from 10 to 20%. 
Their breaking stresses range from 70,000 to 150,000 lbs. 
or more per sq. inch. Cast iron, being comparatively brit- 
tle, reaches at rupture an elongation of only 3 or 4 tenths 
of one per cent., the rupturing stress being about 18,000 
lbs. per sq. inch. The elastic limit is rather ill denned in 
the case of this metal ; and the proportion of carbon and 
the mode of manufacture have much influence on its be- 
havior under test. 

193. Lateral Contraction. — In the stretching of prisms of 
nearly all kinds of material, accompanying the elongation 
of length is found also a diminution of width whose rela- 
tive amount in the case of the three metals just treated is 
about i^ or J^ of the relative elongation (within elastic 
limit). Thus, in the third experiment in the table of § 191, 
this relative lateral contraction or decrease of diameter 
= y z to y^ of e, i.e., about 0.00008. In the case of cast 
iron and hard steels contraction is not noticeable ex- 




212 MECHANICS OF ENGINEERING 

cept by very delicate measurements, both within and with- 
out the elastic limit ; but the more ductile metals, as 
wrought iron and the soft steels, when stretched beyond 
the elastic limit show this feature of their deformation 
in a very marked degree. Fig. 201 shows by dotted lines 
the original contour of a wrought iron rod, while the con- 
tinuous lines indicate that at rupture. At the cross section 
of rupture, whose position is determined by some 
local weakness, the drawing out is peculiarly 
pronounced. 

The contraction of area thus produced is some- 
times as great as 50 or 60% at the fracture. 

194. "Flow of Solids." — When the change in re^ 
lative position of the elements of a solid is ex- 
treme, as occurs in the making of lead pipe, 
! drawing of wire, the stretching of a rod of duc- 
tile metal as in the preceding article, we have 
Fig. 201. instances of what is called the Flow of Solids, in- 
teresting experiments on which have been made by 
Tresca. lt -~ . /;.;, : , i 

195. Moduli of Tenacity. — The tensile stress per square 
inch (of original sectional area) required to rupture a 
prism of a given material will be denoted by T and called 
the modulus of ultimate tenacity ; similarly, the modulus of 
safe tenacity, or greatest safe tensile stress on an element, 
by T' ; while the tensile stress at elastic limit may be 
called T". The ratio of T to T" is not fixed in practice 
but depends upon circumstances (from y^ to 2 /^). 

Hence, if a prism of any material sustains a total pull 
or load P, and has a sectional area=i^, we have 

P=FT for the ultimate or breaking load. \ 

P=FT' " " safe load. V . . (2) 

P'=FT" " " load at elastic limit. ) 

Of course 3T' -should always be less than T '". 



TENSION. 213 

196. Resilience of a Stretched Prism. — Fig. 202. In the 
gradual stretching of a prism, fixed at one extremity, the 
value of the tensile force P at the other necessarily de- 
pends on the elongation X at each stage of the lengthening, 
according to the relation [eq. (1) of § 191.] 



■ . . ' . (3) 



FE, 



within the elastic limit. (If we place a weight G on the 
.r-^ flanges of the unstretched prism and then leave 
— it to the action of gravity and the elastic action 
of the prism, the weight begins to sink, meeting 
an increasing pressure P, proportional to X, from 
the flanges). Suppose the stretching to continue 
until P reaches some value P" (at elastic limit 
\\ say), and X a value X". Then the work done so 

fig. 202. &= mean force X space = }4 P" X" . . (4) 
But from (2) P'=FT", and (see §§ 184 and 191) 

X"=z"l 
.-. (4) becomes U^yi T" e". Fl=% T" e" V . . (5) 

where Fis the volume of the prism. The quantity }iT"e", 
or work done in stretching to the elastic limit a cubic 
inch (or other unit of volume) of the given material, Weis- 
bach calls the Modulus of Resilience for tension. From (5) 
it appears that the amounts of work done in stretching to 
the elastic limit prisms of the same material but of differ- 
ent dimensions are proportional to their volumes simply. 
The quantity }4T"e" is graphically represented by the 
area of one of the triangles such as OA'B, OA'B" in Fig. 
200 ; for (in the curve for wrought iron for instance) the 
modulus of tenacity at elastic limit is represented by A'B, 
and e" (i.e., e for elastic limit) by OA. The remainder of 



214 MECHANICS OF ENGINEERING. 

the area OBG included between the curve and the hori- 
zontal axis, i.e., from B to (7, represents the work done in 
stretching a cubic unit from the elastic limit to the point 
of rupture, for each vertical strip having an altitude =p 
and a width =de, has an area =pde, i.e., the work done by 
the stress p on one face of a cubic unit through the dis- 
tance ds, or increment of elongation. 

If a weight or load = G be " suddenly "applied to stretch 
the prism, i.e., placed on the flanges, barely touching 
them, and then allowed to fall, when it comes to rest again 
it has fallen through a height X lt and experiences at this 
instant some pressure P x from the flanges; Pi=? The 
work GX\ has been entirely expended in stretching the 
prism, none in changing the kinetic energy of G, which 
=0 at both beginning and end of the distance ^, 

••• G^=y 2 P^ .-. P,= 2£. 

Since Pr=26r, i.e., is > G, the weight does not remain in 
this position but is pulled upward by the elasticity of the 
prism. In fact, the motion is harmonic (see §§ 59 and 
138). Theoretically, the elastic limit not being passed, the 
oscillations should continue indefinitely. 

Hence a load G " suddenly applied " occasions double the 
tension it would if compelled to sink gradually by a sup- 
port underneath, which is not removed until the tension is 
just = G, oscillation being thus prevented. 

If the weight G sinks through a height —h before strik- 
ing the flanges, Fig. 202, we shall have similarly, within 
elastic limit, if X x = greatest elongation, (the mass of rod 
being small compared with that of G). 

G(h+h)=y*PA .... (6) 

If the elastic limit is to be just reached we have from eqs. 
(5) and (6), neglecting ^ compared with h, 

Gh=y 2 T"s"V ... (7) 



TENSION. 



215 



an equation of condition that the prism shall not be in- 
jured. 

Example. — If a steel prism have a sectional area of y£ 
sq. inch and a length 1=10 ft. =120 inches, what is the 
greatest allowable height of fall of a weight of 200 lbs., 
that the final tensile stress induced may not exceed T"= 
30,000 lbs. per sq. inch, if e" =.002 ? From (7), using the 
inch and pound, we have 



h= 



rVF_30,000x.002x^xl20 =45 incheg> 



2G 



2x200 



dx 



197. Stretching of a Prism by Its Own Weight. — In the case 
of a very long prism such as a mining- 
pump rod, its weight must be taken into 
account as well as that of the terminal 
load P„ see Fig. 203. At (a.) the prism 
is shown in its unstrained condition ; at 
(b) strained by the load P x and its own 
weight. Let the cross section be =F, the 
heaviness of the prism =y. Then the rela- 
tive extension of any element at a distance 



t 

o 

(a: 



T~ 



J(*J 



Fig. 203. 



x from o is 



dl (P x + r Fx) 



dx' 



FE t 



(1) 



(See eq. (1) § 191) ; since P v -\-Fyx is the load hanging upon 
the cross section at that locality. Equal dx's, therefore, 
are unequally elongated, x varying from to I. The total 
elongation is 



x= rax 



-i 



L • C l [P.dxi-rFxdx] = ? il 

do 



Gl 



FE { 



FE t FE t 



I.e., /= the amount due to P lt plus an extension which 
half the weight of the prism would produce, hung at the 
lower extremitv. 



216 MECHANICS OF ENGINEERING. 

Tke foregoing relates to the deformation of the piece, 
and is therefore a problem of stiffness. As to the strength 
of the prism, the relative elongation £=dX-r-dx [see eq. (1)], 
which is variable, must nowhere exceed a safe value e'= 
T'~E L (from eq. (1) § 191, putting P=FT f , and X=Z). 
Now the greatest value of the ratio dXiclx, by inspecting 
eq. (1), is seen to be at the upper end where x—l. The 
proper cross section F, for a given load P lt is thus found. 

Putting Isp^J^, we have F =_|i_ . (2) 

198. Solid of Uniform Strength in Tension, or hanging body 
of minimum material supporting its oivn 
iveight and a terminal load P x . Let it be a 
solid of revolution. If every cross-section 
F at a distance =x from the lower extrem- 
ity, bears its safe load FT', every element 
of the body is doing full duty, and its form 
is the most economical of material. 

The lowest section must have an area 

fig. 204. F Q =P X + T \ since A is its safe load. Fig. 

204. Consider any horizontal lamina ; its weight is yFdx, 

(j= heaviness of the material, supposed homogenous), and 

its lower base F must have P x + G for its safe load, i.e. 

G+P^FT' ... (1) 

in which G denotes the weight of the portion of the solid 
below F. Similarly for the upper base F-{-dF, we have 

G+P l + r Fdx=(F+dF)T' . . (2) 

By subtraction we obtain 

r Fdx=T'dF; i.e. -j=dx= ^ 
T F 




TENSION. 217 

in which the two variables x and F are separated. By in- 
tegration we now have 






F, 



(3) 



\x P, yx 

i.e., F=FtfF =^r er (4) 

from which .Fmay be computed for any value of x. 

The weight of the portion below any F is found from (1) 
and (4) ; i.e. 

g=p 1 {3—i)\ (5) 

while the total extension X will be 

i= *'%n l < 6 ) 

the relative elongation dl-^-dx being the same for every dx 
and bearing the same ratio to e" (at elastic limit), as T' 
does to T". 

199. Tensile Stresses Induced by Temperature. — If the two 

ends of a prism are immovably fixed, when under no strain 
and at a temperature t, and the temperature is then low- 
ered to a value t\ the body suffers a tension proportional 
to the fall in temperature (within elastic limit). If for a 
rise or fall of 1° Fahr. (or Cent.) a unit of length of the 
material would change in length by an amount tj (called 
the co-efficient of expansion) a length =1 would be con- 
tracted an amount X=rjl(t-t') during the given fall of tem- 
perature if one end were free. Hence, if this contraction 
is prevented by fixing both ends, the rod must be under a 
tension P, equal in value to the force which would be 



218 MECHANICS OF ENGINEERING. 

necessary to produce the elongation A, just stated, under 
ordinary circumstances at the lower temperature. 

From eq. (1) §191, therefore, we have for this tension 
due to fall of temperature 

P= ^rjl(t-t')=E t F(t—t')y 

For 1° Cent, we may write 

For Cast iron rj = .0000111 ; 
" Wrought iron = .0000120 ; 
" Steel = .0000108 to .0000114 ; 

" Copper 7i = .0000172 ; 

" Zinc 7] = .0000300. 



COMPRESSION OF SHORT BLOCKS. 

200. Short and Long Columns. — In a prism in tension, its 
own weight being neglected, all the elements between the 
localities of application of the pair of external forces pro- 
ducing the stretching are in a state of stress, if the exter- 
nal forces act axially (excepting the few elements in the 
immediate neighborhood of the forces ; these suffering 
local stresses dependent on the manner of application of 
the external forces), and the prism may be of any length 
without vitiating this statement. But if the two external 
forces are directed toward each other the intervening ele- 
ments will not all be in the same state of compressive 
stress unless the prism is comparatively short (or unless 
numerous points of lateral support are provided). A long 
prism will buckle out sideways, thus even inducing tensile 
stress, in some cases, in the elements on the convex side. 

Hence the distinction between short blocks and long 
columns. Under compression the former yield by crush- 
ing or splitting, while the latter give way by flexure (i.e. 
bending). Long columns, then will be treated separately 



COMPRESSION OF SHORT BLOCKS. 219 

in a subsequent chapter. In the present section the blocks 
treated being about three or four times as long as wide, 
all the elements will be considered as being under equal 
compressive stresses at the same time. 

201. Notation for Compression. — By using a subscript c, 
we may write 

U c = Modulus of Elasticity;* i.e. the quotient of the 
compressive stress per unit of area divided by the relative 
shortening ; also 

C= Modulus of crushing; i.e. the force per unit of sec- 
tional area necessary to rupture the block by crushing ; 

C'= Modulus of safe compression, a safe compressive 
stress per unit of area; and 

G"= Modulus of compression at elastic limit. , 

For the absolute and relative shortening in length we 
may still use X and e, respectively, and within the elastic 
limit may write equations similar to those for tension, F 
being the sectional area of the block and P one of the ter- 
minal forces, while p = compressive stress per unit of area 
of F, viz.: 

W-P- P+F _ P+F _Pl m 

c e dX-~-dx X+l FX '_•••• W 

within the elastic limit. 
Also for a short block 

Crushing force =FC \ 

Compressive force at elastic limit —FC" > . (2) 
Safe compressive force =FC ) 

202. Remarks on Crushing. — As in § 182 for a tensile 
stress, so for a compressive stress we may prove th-at a 

* [Note. — It must be remembered that the modulus of elasticity, 
whether for normal or shearing stresses, is a number indicative of stiff- 
ness, not of strength, and has no relation to the elastic limit (except 
that experiments to determine it must not pass that limit).] 



220 MECHANICS OF ENGINEERING. 

shearing stress =p sin a cos a is produced on planes at an 
angle a with the axis of the short block, p being the com- 
pression per unit of area of transverse section. Accord- 
ingly it is found that short blocks of many comparatively 
brittle materials yield by shearing on planes making an 
angle of about 45° with the axis, the expression p sin a 
cos a reaching a maximum, for «=45° ; that is, wedge- 
shaped pieces are forced out from the sides. Hence the 
necessity of making the block three or four times as long 
as wide, since otherwise the friction on the ends would 
cause the piece to show a greater resistance by hindering 
this lateral motion. Crushing by splitting into pieces 
parallel to the axis sometimes occurs. 

Blocks of ductile material, however, yield by swelling 
out, or bulging, laterally, resembling plastic bodies some- 
what in this respect. 

The elastic limit is more difficult to locate than in ten- 
sion, but seems to have a position corresponding to that 
in tension, in the case of wrought iron and steel. With 
cast iron, however, the relative compression at elastic 
limit is about double the relative extension (at elastic 
limit in tension), but the force producing it is also double. 
For all three metals it is found that E c =E t quite nearly, 
so that the single symbol E may be used for both. 



EXAMPLES IN TENSION AND COMPRESSION. 

203. Tables for Tension and Compression. — The round num- 
bers of the following tables are to be taken as rude averages 
only, for use in the numerical examples following. (The 
scope and design of the present work admit of nothing 
more. For abundant detail of the results of the more im- 
portant experiments of late years, the student is referred 
to the recent works of Profs. Thurston, Burr, Lanza, and 
Wood). Another column might have been added giving 
the Modulus of Kesilience in each case, viz.: V 2 s"T" 



jT" 2 \ 

(which also =-- ~ ) ; see § 196. e is an abstract 



num- 



EXAMPLES IN TENSION AND COMPRESSION. 



221 



ber, and =X-hl, while E u T", and T are given in pounds 
per square inch: 

TABLE OF THE MODULI, ETC., OF MATERIALS IN TENSION. 





e" 


£ 


E t 


rpii 


T 


Material. 


(Elastic limit.) 


At Rupture. 


Mod. of Elast. 


Elastic limit. 


Rupture. 




abst. number. 


abst. number. 


lbs. per sq. in. 


lbs. per sq. in. 


lbs. per sq. in. 


Soft Steel, 


.00200 


.2500 


26,000,000 


50,000 


80,000 


Hard Steel, 


.00200 


.0500 


40,000,000 


90,000 


130,000 


Cast Iron, 


.00066 


.0020 


14,000,000 


9,000 


18,000 
( 45,000 
X to 
( 60,000 

16,000 
to 

50,000 


Wro't Iron, 
Brass, 


.00080 
.00100 


.2500 


28,000,000 
10,000,000 


22,000 

( 7,000 
\ to 
( 19,000 


Glass, 






9,000,000 




3,500 


Wood, with 
the fibres, 


( .00200 
•< to 
( .01100 


.0070 

to 
.0150 


200,000 
to 
2,000,000 


3,000 

to 
19,000 


6,000 

to 
28,000 


Hemp rope, 










7,000 



[N.B.— Expressed in kilograms per square centim., E t , T and T" would be nu 
merically about 1 / ]4 as large as above, while e and e" would be unchanged.] 



TABLE OF MODULI, ETC.; COMPRESSION OF SHORT BLOCKS. 





1! 

£ 


£ 


E 


C" 


C 


Material. 


Elastic limit. 


At Rupture. 


Mod. of Elast. 


Elastic limit. 


Rupture. 




abst. number 


abst. number. 


lbs. per sq. in. 


lbs. per sq. in. 


lbs. per sq. in. 


Soft Steel, 


0.00100 




30,000,000 


30,000 




Hard Steel. 


0.00120 


0.3000 


40,000,000 


50,000 


200,000 


Cast Iron, 


0.00150 




14,000,000 


20,000 


90,000 


Wro't Iron, 


0-00080 


0.3000 


28,000,000 


24,000 


40,000 


Glass, 










20,000 


Granite, 










10,000 


Sandstone, 










5,000 


Brick, 










3,000 


Wood, with 
the fibres, 




( 0.0100 

•< to • 
( 0.0400 


350,000 

to 

2,000,000 




2,000 

to 
10,000 


Portland I 
Cement, f 










4,000 



222 MECHANICS OF ENGINEERING. 

204. Examples. No. 1. — A bar of tool steel, of sectional 
area =0.097 sq. inches, is ruptured by a tensile force of 
14,000 lbs. A portion of its length, originally y 2 a foot, 
is now found to have a length of 0.532 ft. Required T, 
and £ at rupture. Using the inch and pound as units (as 
in the foregoing tables) we have 2=i^°=144326 lbs. per 
sq. in.; (eq. (2) § 195); while 

£=(0.532— 0.5) x 12^(0.50 x 12) =0.064. 

Example 2. — Tensile test of a bar of " Hay Steel " for 
the Glasgow Bridge, Missouri. The portion measured was 
originally 3.21 ft. long and 2.09 in. X 1.10 in. in section. 
At the elastic limit P was 124,200 lbs., and the elongation 
was 0.064 ins. Required U t , T', and e" (for elastic limit). 

e"=* = ^ 06 t =.00165 at elastic limit. 
I 3.21x12 

T"=124,200--(2.09x 1.10)= 54,000 lbs. per sq. in. 

E ^=wr^kr^ m ' m lbs - * er •* in - 

Nearly the same result for E t would probably have been 
obtained for values of p and £ below the elastic limit. 

The Modulus of Resilience of the above steel (see § 196) 
would be ]/ 2 e" T"= 44.82 inch -pounds of work per cubic 
inch of metal, so that the whole work expended in stretch- 
ing to the elastic limit the portion above cited is 

U= % e" T" T=3968. inch -lbs. 

An equal amount of work will be done by the rod in re- 
covering its original length. ~- im 

Example 3. — A hard steel rod of y& sq. in. section and 
20 ft. long is under no stress at a temperature of 130° 



EXAMPLES IN TENSION AND COMPRESSION. 223 

Cent., and is provided with flanges so that the slightest 
contraction of length will tend to bring two walls nearer 
together. If the resistance to this motion is 10 tons how 
low must the temperature fall to cause any motion ? y be- 
ing = .0000120 (Cent, scale). From § 199 we have, ex- 
pressing P in lbs. and F in sq. inches, since E = 40,000,000 
lbs. per sq. inch, 

10x2,000=40,000,000 x % X (130-*') X 0.000012; whence 
£'=46.6° Centigrade. 

Example 4. — If the ends of an iron beam bearing 5 tons 
at its middle rest upon stone piers, required the necessary 
bearing surface at each pier, putting C for stone =200 
lbs. per sq. inch. 25 sq. in., Ans. 

Example 5. — How long must a wrought iron rod be, 
supported vertically at its upper end, to break with its 
own weight ? 216,000 inches, Ans. 

Example 6. — One voussoir (or block) of an arch -ring 
presses its neighbor with a force of 50 tons, the joint hav- 
ing a surface of 5 sq. feet ; required the compression per 
sq. inch. 138.8 lbs. per sq. in., Ans. 

205. Factor of Safety. — When, as in the case of stone, the 
value of the stress at the elastic limit is of very uncertain 
determination by experiment, it is customary to refer the 
value of the safe stress to that of the ultimate by making 
it the %'th portion of the latter, n is called a factor of 
safety, and should be taken large enough to make the safe 
stress come within the elastic limit. For stone, n should 
not be less than 10, i.e. C' = C+n; (see Ex. 6, just given). 



206. Practical Notes. — It was discovered independently by 
Commander Beardslee and Prof. Thurston, in 1873, that 
if wrought iron rods were strained considerably beyond 
the elastic limit and allowed to remain free from stress 



224 MECHANICS OF ENGINEERING. 

for at least one day thereafter, a second test would show 
higher limits both elastic and ultimate. 

When articles of cast iron are imbedded in oxide of iron 
and subjected to a red heat for some days, the metal loses 
most of its carbon, and is thus nearly converted into 
wrought iron, lacking, however, the property of welding. 
Being malleable, it is called malleable cast iron. 

Chrome steel (iron and chromium) and tungsten steel pos- 
sess peculiar hardness, fitting them for cutting tools, rock 
drills, picks, etc. 

B y fatigue of metals we understand the fact, recently dis- 
covered by Wohler in experiments made for the Prussian 
Government, that rupture may be produced by causing the 
stress on the elements to vary repeatedly between two 
limiting values, the highest of which may be considerably 
below T (or (7), the number of repetitions necessary to 
produce rupture being dependent both on the range of 
variation and the higher value. 

For example, in the case of Phoenix iron in tension, 
rupture was produced by causing the stress to vary from 
to 52,800 lbs. per sq. inch, 800 times ; also, from to 
44,000 lbs. per sq. inch 240,853 times ; while 4,000,000 va- 
riations between 26,400 and 48,400 per sq. inch did not 
cause rupture. Many other experiments were made and 
the following conclusions drawn (among others): 

Unlimited repetitions of variations of stress (lbs. per 
sq. in.) between the limits given below will not injure the 
metal (Prof. Burr's Materials of Engineering). 

Wrought iron i From 17 > 600 Com P* to 17 ' 600 Tension ' 
g " ( " to 33,000 

( From 30,800 Comp. to 30,800 Tension. 

Axle Cast Steel. ■] " to 52,800 

( " 38500 Tens, to 88,000 



SHEARING. 

SHEARING. 



225 



207. Rivets. — The angular distortion called shearing 
strain in the elements of a body, is specially to be provided 
for in the case of rivets joining two or more plates. This 
distortion is shown, in Figs. 205 and 206, in the elements 
near the plane of contact of the plates, much exaggerated. 



3ULZJ-T* 



Fig. 205. 



3 — 



-f€ 



Fig. 206. 



In Fig. 205 (a lap-joint) the rivet is said to be in single 
shear ; in Fig. 206 in double shear. If P is just great 
enough to shear off the rivet, the modulus of ultimate shear- 
ing, which may be called S, (being the shearing force per 
unit of section when rupture occurs) is 



F %7id 2 



(1) 



in which F= the cross section of the rivet, its diameter 
being =d. For safety a value S'= % to yi of S should 
be taken for metal, in order to be within the elastic limit. 

As the width of the plate is diminished by the rivet 
hole the remaining sectional area of the plate should be 
ample to sustain the tension P, or 2P, (according to the 
plate considered, see Fig. 206), P being the safe shearing 
force for the rivet. Also the thickness t of the plate 
should be such that the side of the hole shall be secure 
against crushing ; P must not be > C'td, Fig. 205. 

Again, the distance a, Fig. 205, should be such as to 
prevent the tearing or shearing out of the part of the 
plate between the rivet and edge of the plate. 



226 



MECHANICS OF ENGINEERING. 



For economy of material the seam or joint should be 
no more liable to rupture by one than by another, of the 




Fig. 207. 

four modes just mentioned. The relations which must 
then subsist will be illustrated in the case of the " butt- 
joint " with two cover-plates, Fig. 207. Let the dimen- 
sions be denoted as in the figure and the total tensile force 
on the joint be = Q. Each rivet (see also Fig. 206) is ex- 
posed in each of two of its sections to a shear of } 2 Q f 
hence for safety against shearing of rivets we put 



Q=%t:cPS- 



(1) 



Along one row of rivets in the main plate the sectional 
area for resisting tension is reduced to (b — 3d)t lf hence for 
safety against rupture of that plate by the tension Q, we 
put 



Q^b—ZdW 



(2) 



Equations (1) and (2) suffice to determine d for the rivets 
and t x for the main plates, Q and b being given; but the 
values thus obtained should also be examined with refer- 
ence to the compression in the side of the rivet hole, i.e., 
y 6 Q must not be > C't^d. [The distance a, Fig. 205, to the 
edge of the plate is recommended by different authorities 
to be from d to 3d.] 

Similarly, for the cover-plate we must have 



and } 2 Q not >C'td. 



y 2 Qor(b- 

< 



-3d)tT' 



(3) 



SHEARING. 227 

If the rivets do not fit their holes closely, a large margin 
should be allowed in practice. Again, in boiler work, the 
pitch, or distance between centers of two consecutive rivets 
may need to be smaller, to make the joint steam-tight, than 
would be required for strength alone, 

208. Shearing Distortion. — The change of form in an ele- 
ment due to shearing is an angular deformation and will 
be measured in tt -measure. This angular change or dif- 
ference between the value of the corner angle during strain 
and )4n> its value before strain, will be called o, and is 
proportional (within elastic limit) to the shearing stress 
per unit of area, p s , existing on all the four faces whose 
angles with each other have been changed. 

Fig. 208. (See § 181). By § 184 the Modulus of Shearing 
Elasticity is the quotient obtained by dividing p s by d ; i.e.. 
(elastic limit not passed), 

K=f . . . . (l) 

or inversely, o=p s ~-E s (1)' 

The value of F s for different substances is most easily 
determined by experiments on torsion 
in which shearing is the most promi- 
nent stress. (This prominence depends 
on the position of the bounding planes, 
of the element considered ; e.g., in Fig. 
208, if another element were considered 
within the one there shown and with 
fig. 208. its planes at 45° with those of the first, 

we should find tension alone on one pair of opposite faces, 
compression alone on the other pair.) It will be noticed 
that shearing stress cannot be present on two opposite 
faces only, but exists also on another pair of faces (those 
perpendicular to the stress on the first), forming a couple 
of equal and opposite moment to the first, this being 
necessary for the equilibrium of the element, even when 




228 



MECHANICS OF ENGINEERING. 



tensile or compressive stresses are also present on the 
faces considered. 

209. Shearing Stress is Always of the Same Intensity on the 
Four Faces of an Element. — (By intensity is meant per unit 
of area ; and the four faces referred to are those perpen- 
dicular to the paper in Fig. 208, the shearing stress being 
parallel to the paper.) 

Let dx and dz be the width and height of the element 
in Fig. 208, while dy is its thickness perpendicular to the 
paper. Let the intensity of the shear on the right hand 
face be =q^ that on the top face =p s . Then for the ele- 
ment as a free body, taking moments about the axis per- 
pendicular to paper, we have 

q s dz dy X dx — p s dx dy x dz =0 .*. q s =p s 
(dx and dz being the respective lever arms of the forces 
g s dz dy and p s dx dy.) 

Even if there were also tensions (or compressions) on 
one or both pairs of faces their moments about would 
balance (or fail to do so by a differential of a higher order) 
independently of the shears, and the above result would 
still hold. 



210. Table of Moduli for Shearing. 





8" 


& 


s« 


s 


-? 

Material. 


i.e. S at elastic 
limit. 


Mod. of Elasticity 
for Shearing. 


(Elastic limit.) 


(Rupture.) 




arc in jr-measure. 


lbs. per sq. in. 


lbs. per sq. in 


lbs. per sq. in. 


Soft Steel, 




9,000,000 




70,000 


Hard Steel, 


0.0032 


14,000,000 


45,000 


90,000 


Cast Iron, 


0.0021 


7,000,000 


15,000 


30,000 


Wrought Iron, 


0.0022 


9,000,000 


20,000 


50,000 


Brass, 




5,000,000 






Glass, 










Wood, across ( 
fibre, | 








1,500 
to 

8,000 


Wood, along ( 








500 
to 

1 .-200 


fibre, ( 









SHEARING. 229 

As in the tables for tension and compression, the above 
values are averages. The true values may differ from 
these as much as 30 per cent, in particular cases, accord- 
ing to the quality of the specimen. 

211. Punching rivet holes in plates of metal requires the 
overcoming of the shearing resistance along the convex 
surface of the cylinder punched out. Hence if d = diam- 
eter of hole, and t = the thickness of the plate, the neces- 
sary force for the punching, the surface sheared being 
F= tnd, is 

P=St7rd .... (2) 

Another example of shearing action is the " stripping '* 
of the threads of a screw, when the nut is forced off lon- 
gitudinally without turning, and resembles punching in 
its nature. 

212. E and E g • Theoretical Relation. — In case a rod is in 
tension within the elastic limit, the relative (linear) lateral 
contraction (let this =m) is so connected with E t and E s 
that if two of the three are known the third can be de- 
duced theoretically. This relation is proved as follows, 
by Prof. Burr. Taking an elemental cube with four of its 
faces at 45° with the axis of the piece, Fig. 209, the axial 
half-diagonal AD becomes of a length AD'=AD+e.AD 
under stress, while the transverse half diagonal contracts 
to a length B / D / =AD—m.AD. The angular distortion 3 





A?** . 


A< 






Fig. 200. § 212. 




Fig. 210. 



230 MECHANICS OF ENGINEERING. 

is supposed very small compared with 90° and is due to 
the shear p s per unit of area on the face BG (or BA). 
From the figure we have 

tan(45°— _) = _.~_.=1— m— e, approx. 

[But, Fig. 210, tan(45° — x)=l — 2x nearly, where x is a 
small angle, for, taking CA=\nAty= AE, tan AD=AF= 
AE—EF. Now approximately EF=WG,^/2^ndiEG^ 
BD^/2 =x^/2 .*. AF= 1— 2x nearly.] Hence 

l_<5=l__ m _ £; or d=m+£ . . (2) 

Eq. (2) holds good whatever the stresses producing the 
deformation, but in the present case of a rod in tension, 
if it is an isotrope, and if p = tension per unit of area on 
its transverse section, (see § 181, putting a=45°), we have 
E t =p+£ and E s =(p s on BC)-±-o= %p^-o. Putting also 
(m :e)=r, whence m=rs, eq. (2) may finally be written 

k Hr+1) h'^'^m ■ ■ (3) 

Prof. Bauschinger, experimenting with cast iron rods, 
found that in tension the ratio m : e was = $), as an average, 
which in eq. (3) gives 

E 6 = — E t =-E t nearly. . . . (4) 

His experiments on the torsion of cast iron rods gave 
^.= 6,000,000 to 7,000,000 lbs. per sq. inch. By (4), then, 
E t should be 15,000,000 to 17,500,000 which is approxi- 
mately true (§ 203). 

Corresponding results may be obtained for short blocks 
in compression, the lateral change being a dilatation in- 
stead of a contraction. 




SHEARING. 231 

213. Examples in Shearing. — Example 1. — Kequired the 
proper length, a, Fig. 211, to 
guard against the shearing off, 
along the grain, of the portion 
ab, of a wooden tie-rod, the force 
P being = 2 tons, and the width 
of the tie = 4 inches. Using a 
value of S' = 100 lbs. per sq. in., 
we put &*#'= 4,000 cos 45° ; i.e. 
fio. 211. a= (4,000x0.707) --(4x100)= 7.07 

inches. 

Example 2. — A i/% in. rivet of wrought iron, in single 
shear (see Fig, 205) has an ultimate shearing strength 
P= FS=%7zd 2 S= %7i(7/ 8 ) 2 x 50,000= 30,050 lbs. For safety, 
putting #'=8,000 instead of #,P =4,800 lbs. is its safe 
shearing strength in single shear. 

The wrought iron plate, to be secure against the side- 
crushing in the hole, should have a thickness t, computed 
thus : 

P'=tdO\ or 4,800=^x12,000 .\ £=0.46 in. 

If the plate were only 0.23 in. thick the safe value of P 
would be only % of 4,800. 

Example 3. — Conversely, given a lap-joint, Fig. 205, in 
which the plates are y^ in. thick and the tensile force on 
the joint = 600 lbs. per linear inch of seam, how closely 
must ^ inch rivets be spaced in one row, putting £'=8,000 
and C =12,000 lbs. per sq. in. ? Let the distance between 
centres of rivets be =x (in inches), then the force upon 
each rivet =600#, while its section i^=0.44sq. in. Having 
regard to the shearing strength of the rivet we put 600x= 
0.44x8,000 and obtain cc=5.86 in.; but considering that the 
safe crushing resistance of the hole is = ^-^.12,000= 
2,250 lbs., 600^=2,250 gives ^=3.75 inches, which is the 
pitch to be adopted. What is the tensile strength of the 
reduced sectional area of the plate, with this pitch ? 



232 MECHANICS OF ENGINEERING. 

Example 4. — Double butt-joint ; (see Fig. 207) ; 3 / 8 inch 
plate; ^ in. rivets; T'= G'=12,000 ; #'=8,333; width of 
plates=14 inches. Will one row of rivets be sufficient at 
each side of joint, if $=30,000 lbs.? The number of rivets 
= ? Here each rivet is in double shear and has therefore 
a double strength as regards shear. In double shear the 
safe strength of each rivet =2i^S"=7,333 lbs. Now 30,000^ 
7,333=40 (say). "With the four rivets in one row the re- 
duced sectional area of the main plate is =[14 — 4x ^] X 3 / 8 
=4.12 sq. in., whose safe tensile strength is =i^T =4.12 X 
12,000=49,440 lbs.; which is > 30,000 lbs. .-. main plate is 
safe in this respect. But as to side-crushing in holes 
in main plate we find that G%d (i.e. 12,000 X 3 / 8 X ^=3,375 
lbs.) is <%Q i.e.<7,500 lbs., the actual force on side of 
hole. Hence four rivets in one row are too few unless 
thickness of main plate be doubled. Will eight in one 
row be safe ? 



TORSION. 



233 



CHAPTEE II. 

TORSION. 

214. Angle of Torsion and of Helix. When a cylindrical 
beam or shaft is subjected to a twisting or torsional action, 
i. e. when it is the means of holding in equilibrium two 
couples in parallel planes and of equal and opposite mo- 
ments, the longitudinal axis of symmetry remains straight 

and the elements along it exper- 
ience no stress (whence it may be 
Icalled the "line of no twist"), 
while the lines originally parallel to 
it assume the form of helices, each 
distorted in its angles (originally 




is 



Fig. 212. 

element of which 
right angles), the amount of distortion being assumed pro- 
portional to the radius of the helix. The directions of the 




faces of any element were originally as follows : two radial, 
two in consecutive transverse sections, and the other two 
tangent to two consecutive circular cylinders whose com- 
mon axis is that of the shaft. E.g. in Fig. 212 we have 
an unstrained shaft, while in Fig. 213 it holds the two 



234 



MECHANICS OF ENGINEERING. 



couples (of equal moment P a = Q b) in equilibrium. These 
couples act in parallel planes perpendicular to the axis of 
the prism and a distance, l 9 apart. Assuming that the 
transverse sections remain plane and parallel during tor- 
sion, any surface element, m, which in Fig. 212 was entire- 
ly right-angled, is now distorted. Two of its angles have 
been increased, two diminished, by an amount d, the angle 
between the helix and a line parallel to the axis. Suppos- 
ing m to be the most distant of any element from the axis, 
this distance being e, any other element at a distance z 

from the axis experiences an angular distortion =- d. 



If now we draw B' parallel to 0' A the angle B B', 
=«, is called the Angle of Torsion, while d may be called the 
helix angle; the former lies in a transverse plane, the latter 
in a plane tangent to the cylinder. Now 



tan d =(linear arc B B')-7-l; but lin. arc B B' 
putting d for tan d, (d being small) 



ea ; hence, 



eot 
"T 



(1) 



(S and a both in n measure). 

215. Shearing Stress on the Elements. The angular distor- 
tion, or shearing strain, d, of any element (bounded as al- 
ready described) is due to the shearing stresses exerted on 
it by its neighbors on the four faces perpendicular to the 

tangent plane of the cylindri- 
cal shell in which the element 
is situated. Consider these 
neighboring elements of an 
outside element removed, and 
the stresses put in ; the latter 
are accountable for the dis- 
riG> 214 " tortion of the element and so 




TOKSIOX. 2o5 

hold it in equilibrium. Fig. 214 shows this element 
" free." Within the elastic limit d is known to be propor- 
tional to p s , the shearing stress per unit of area on the 
faces whose relative angular positions have been changed. 
That is, from eq. (1) § 208, d=p^E s ; whence, see (1) of 
§ 214, 

*-^* W 

In (2) p s and e both refer to a surface element, e being 
the radius of the cylinder, and p s the greatest intensity of 
shearing stress existing in the shaft. Elements lying nearer 
the axis suffer shearing stresses of less intensity in pro- 
portion to their radial distances, i.e., to their helix-angles. 
That is, the shearing stress on that face of the element 
which forms a part of a transverse section and whose dis- 
tance from the axis is z, is p } =— p s , per unit of area, and 

e 

the total shear on the face is pdF, dF being the area of the 
face. 



216. Torsional Strength. — We are now ready to expose the 
full transverse section of a shaft under torsion, to deduce 
formulae of practical utility. Making a right section of 
the shaft of Fig. 213 anywhere between the two couples 
and considering the left hand portion as a free body, the 
forces holding it in equilibrium are the two forces P of 
the left-hand couple and an infinite number of shearing 
forces, each tangent to its circle of radius z, on the cross 
section exposed by the removal of the right-hand portion. 
The cross section is assumed to remain plane during tor- 
sion, and is composed of an infinite number of dF's, each 
being the area of an exposed face of an element ; see Fig, 
215. 



236 



MECHANICS OF ENGINEERING. 

Aw* 




Each elementary shearing force = -± p^dF, and z is its 
lever arm about the axis Oo . Eor equilibrium, I (mom.) 
about the axis Oo must =0 ; i.e. in detail 

—Py 2 a—Py 2 a+ J ( S p H dF)z=0 



or, reducing, 



^ s Cz 2 dF=Pa ; or, 



PJi 



*L=Pa 



(3) 



Eq. (3) relates to torsional strength, since it contains p s , the 
greatest shearing stress induced by the torsional couple, 
whose moment Pa is called the Moment of Torsion, the 
stresses in the cross section forming a couple of equal and 
opposite moment. 

7 P is recognized as the Polar Moment of Inertia of the cross 
section, discussed in § 94 ; e is the radial distance of the 
outermost element, and = the radius for a circular shaft. 

217. Torsional Stiffness. — In problems involving the angle 
of torsion, or deformation of the shaft, we need an equa- 
tion connecting Pa and a, which is obtained by substitut- 
ing in eq. (3) the value of p s in eq. (2), whence 



a I^=Pa 



(4) 



toksiok. 237 

Prom this it appears that the angle of torsion, «, is propor- 
tional to the moment of torsion, Pa, within the elastic 
limit ; a must be expressed in it-measure. Trautwine cites 1° 
(i.e. a= 0.0174) as a maximum allowable value for shafts. 

218. Torsional Resilience is the work done in twisting a 
shaft from an unstrained state until the elastic limit is 
reached in the outermost elements. If in Fig. 213 we 
imagine the right-hand extremity to be fixed, while the 
other end is gradually twisted through an angle «i each 
force P of the couple must be made to increase gradually 
from a zero value up to the value P l9 corresponding to a { . 
In this motion' each end of the arm a describes a space 
= }4aa lf and the mean value of the force = }4P\ (compare 
§ 196). Hence the work done in twisting is 

Uy=%P l xy 2 aa l X i l=y 2 P l aa l . . (5) 
By the aid of preceding equations, (5) can be written 

If for p s we write S' (Modulus of safe shearing) we have 
for the safe resilience of the shaft 

l7W ££ • • • • (7) 

If the torsional elasticity of an originally unstrained shaft 
is to be the means of arresting the motion of a moving 
mass whose weight is G, (large compared with the parts 
intervening) and velocity =v, we write (§ 133) 

9 2 
as the condition that the shaft shall not be injurecL 



238 MECHANICS OF ENGINEERING. 

219. Polar Moment of Inertia. — For a shaft of circular 
cross section (see § 94) ip=^7zr 4 ; for a hollow cylinder 
7p == ^^(r 1 4 — r 2 4 ) ; while for a square shaft 7 p =i^6 4 , b being 
the side of the square ; for a rectangular cross-section 
sides b and h, I p =i 2 hh(¥-\-h 2 ). For a cylinder e=r; if hol- 
low, e=r , the greater radius. For a square, e=}^b^/2. 

220. Non-Circular Shafts. — If the cross-section is not cir- 
cular it becomes warped, in torsion, instead of remaining 
plane. Hence the foregoing theory does not strictly ap- 
ply. The celebrated investigations of St. Venant, how- 
ever, cover many of these cases. (See § 708 of Thompson 
and Tait's Natural Philosophy ; also, Prof. Burr's Elas- 
ticity and Strength of the Materials of Engineering). His 
results give for a square shaft (instead of the 

- Pa of eq. (4) of § 217), 



61 



Pa=0.841 a¥ f* . . . . (1) 

66 



and Pa=Jl^p s , instead of eq. (3) of § 216, p s being the 
greatest shearing stress. 

The elements under greatest shearing strain are found 
at the middles of the sides, instead of at the corners, when 
the prism is of square or rectangular cross-section. The 
warping of the cross-section in such a case is easily veri- 
fied by the student by twisting a bar of india-rubber in 
his fingers. 

221. Transmission of Power. — Fig. 216. Suppose the cog- 
wheel B to cause A, on the 
same shaft, to revolve uni- 
formly and overcome a resis- 
tance Q, the pressure of the 
teeth of another cog-wheel, 
B being driven by still another 
fig. 216. wheel. The shaft AB is un- 




TORSION. 239 



der torsion, the moment of torsion being = Pa= Qb. (P x 
and Qi the bearing reactions have no moment about the 
axis of the shaft). If the shaft makes u revolutions per 
unit-time, the work transmitted (transmitted ; not expend- 
ed in twisting the shaft whose angle of torsion remains 
constant, corresponding to Pa) per unit-time, i.e. the Power, 
is 

L=P.27ra.u=27TuPa . . . (8) 

To reduce L to Horse Power (§ 132), we divide by JSf, 
the number of units of work per unit-time constituting 
one H. P. in the system of units employed, i.e., 

Horse Power = H. P.= 2;I 



jsr 

For example JV=33,000 ft. -lbs. per minute, or =396,000 
inch -lbs. per minute ; or = 550 ft. -lbs. per second. Usually 
the rate of rotation of a shaft is given in revolutions per 
minute. 

But eq. (8) happens to contain Pa the moment of torsion 
acting to maintain the constant value of the angle of tor- 
sion, and since for safety (see eq. (3) § 216) Pa=/S"/ p -4-e, 
with 7 P = y 2 7ZT^ and e=r for a solid circular shaft, we have 
for such a shaft 

(Safe), fl. P.-2^_ ... (9) 

which is the safe H. P., which the given shaft can trans- 
mit at the given speed. S' may be made 7,000 lbs. per sq. 
inch for wrought iron ; 10,000 for steel, and 5,000 for cast- 
iron. If the value of Pa fluctuates periodically, as when 
a shaft is driven by a connecting rod and crank, for (H. P.) 
we put raX(H. P.), m being the ratio of the maximum to 
the mean torsional moment ; m = about V/ 2 under ordi- 
nary circumstances (Cotterill). 



240 



MECHANICS OF ENGINEERING. 



222. Autographic Testing Machine. — The principle of Prol 
Thurston's invention bearing this name is shown in Fig, 




Fig. 317. 

217. The test-piece is of a standard shape and size, its 
central cylinder being subjected to torsion. A jaw, carry- 
ing a handle (or gear-wheel turned by a worm) and a drum 
on which paper is wrapped, takes a firm hold of one end 
of the test-piece, whose further end lies in another jaw 
rigidly connected with a heavy pendulum carrying a pen- 
cil free to move axially. By a continuous slow motion of 
the handle the pendulum is gradually deviated more and 
more from the vertical, through the intervention of the 
test-piece, which is thus subjected to an increasing tor- 
sional moment. The axis of the test-piece lies in the axis 
of motion. This motion of the pendulum by means of a 
properly curved guide, WR, causes an axial (i.e., parallel 
to axis of test-piece) motion of the pencil A, as well as an 
angular deviation /9 equal to that of the pendulum, and 
this axial distance CF,=sT, of the pencil from its initial 
position measures the moment of torsion =JPa=JPe sin /3. 
As the piece twists, the drum and paper move relatively 
to the pencil through an angle sUo equal to the angle 



TORSION. 



241 



of torsion a so far attained. The abscissa so and ordinate 
sT of the curve thns marked on the paper, measure, 
when the paper is unrolled, the values of a and Pa through 
all the stages of the torsion. Fig. 218 shows typical 




curves thus obtained. Many valuable indications are 
given by these strain diagrams as to homogeneousness of 
composition, ductility, etc., etc. On relaxing the strain 
at any stage within the elastic limit, the pencil retraces 
its path ; but if beyond that limit, a new path is taken 
called an " elasticity -line," in general parallel to the first 
part of the line, and showing the amount of angular re- 
covery, BC, and the permanent angular set, OB. m® MB 

223. Examples in Torsion. — The modulus of safe shearing 
strength, S', as given in § 221, is expressed in pounds per 
square inch ; hence these two units should be adopted 
throughout in any numerical examples where one of the 
above values for S' is used. The same statement applies 
to the modulus of shearing elasticity, U SJ in the table of 
§ 210. rj 

Example 1.— Fig. 216. With P = 1 ton, a = 3 ft., I = 
10 ft., and the radius of the cylindrical shaft r=2.5 inches, 
required the max. shearing stress per sq. inch, p st the 
shaft being of wrought iron. From eq. (3) § 216 



p s 



Pae _ 2,000x36x2.5 



I 



7 2 7rx(2.5) 4 



2,930 lbs. per sq. inch, 



which is a safe value for any ferrous aaetal. 



242 



MECHANICS OF ENGINEERING. 



Example 2. — What H. P. is the shaft in Ex. 1 transmit- 
ting, if it makes 50 revolutions per minute ? Let u = 
number of revolutions per unit of time, and JV= the num- 
ber of units of work per unit of time constituting one 
horse-power. Then H. ~P.=Pu27ra-i-N } which for the foot- 
pound-minute system of units gives 

H. P.=2,000x50x2ttX 3-^33,000=57^ H. P. 

Example 3. — What different radius should be given to 
the shaft in Ex. 1, if two radii at its extremities, originally 
parallel, are to make an angle of 2° when the given moment 
of torsion is acting, the strains in the shaft remaining con- 
stant. From eq. (4) § 217, and the table 210, with a=^ 7:= 
0.035 radians (i.e. n -measure), and I p = 1 / 2 7zr i , we have 



2,000x36x12 

^,t0.035x 9,000,000 



17.45 .-. r=2.04 inches. 



(This would bring about a different p s , but still safe.) The 
foregoing is an example in stiffness. 

Example 4. — A working shaft of steel (solid) is to trans- 
mit 4,000 H. P. and make 60 rev. per minute, the maximum 
twisting moment being \y 2 times the average ; required 
its diameter. <i=14.74 inches. Ans. 

Example 5. — In example 1, p = 2,930 lbs. per square 
inch ; what tensile stress does this imply on a plane at 45° 
with the pair of planes on which p s acts ? Fig. 219 shows 



p t dz 2 



p t dx 2 





dx P. 



Fig. 220. 



TORSION. 24,6 

a small cube, of edge — dx, (taken from the outer helix of 
Fig. 215,) free and in equilibrium, the plane of the paper 
being tangent to the cylinder ; while 220 shows the portion 
BD C, also free, with the unknown total tensile stress pdx 2 ^/2 
acting on the newly exposed rectangle of area =dxxdx^2, 
p being the unknown stress per unit of area. From sym- 
metry the stress on this diagonal plane has no shearing 
component. Putting I [components normal to BD]=0, 
we have 

pdx 2 ,\/2=2dx 2 p !i cos4:5 =dx 2 p s ^/2 ,-.p=p s . (1) 

That is, a normal tensile stress exists in the diagonal 
plane BD of the cubical element equal in intensity to the 
shearing stress on one of the faces, i.e., =2,930 lbs. per sq. 
in. in this case. 

Similarly in the plane AC will be found a compressive 
stress of 2,930 lbs. per sq. in. If a plane surface had been 
exposed making any other angle than 45° with the face of 
the cube in Fig. 219, we should have found shearing and 
normal stresses each less than p s per sq. inch. Hence the 
interior dotted cube in 219, if shown " free " is in tension 
in one direction, in compression in the other, and with 
no shear, these normal stresses having equal intensities. 
Since S' is usually less than T' or C, if p s is made = S r 
the tensile and compressive actions are not injurious. It 
follows therefore that when a cylinder is in torsion any 
helix at an angle of 45° with the axis is a line of tensile, 
or of compressive stress, according as it is a right or left 
handed helix, or vice versa. 

Example 6. — A solid and a hollow cylindrical shaft, of 
equal length, contain the same amount of the same kind 
of metal, the solid one fitting the hollow of the other. 

Compare their torsional strengths, used separately. 
The solid shaft has only ^ the strength of the hollow 
one. Ans. 



244 MECHANICS OE ENGINEERING. 



CHAPTEE III. 

FLEXURE OF HOMOGENEOUS PRISMS UNDER 
PERPENDICULAR FORCES IN ONE PLANE. 

224. Assumptions of the Common Theory of Flexure. — When 
a prism is bent, under the action of external forces per- 
pendicular to it and in the same plane with each other, it 
may be assumed that the longitudinal fibres are in tension 
on the convex side, in compression on the concave side, 
and that the relative stretching or contraction of the ele- 
ments is proportional to their distances from a plane in- 
termediate between, with the understanding that the flex- 
ure is slight and that the elastic limit is not passed in any 
element. 

This " common theory " is sufficiently exact for ordinary 
engineering purposes if the constants employed are prop- 
erly determined by a wide range of experiments, and in- 
volves certain assumptions of as simple a nature as possi- 
ble, consistently with practical facts. These assumptions 
are as follows, (for prisms, and for solids with variable cross 
sections, when the cross sections are similarly situated as 
regards a central straight axis) and are approximately 
borne out by experiment : 

(1.) The externals r " applied " forces are all perpendicu- 
lar to the axis of the piece and lie in one plane, which may 
be called the force-plane ; the force-plane contains the 
axis of the piece and cuts each cross-section symmetri- 
cally ; 

(2.) The cross-sections remain plane surfaces during 
flexure ; 

(3.) There is a surface (or, rather, sheet of elements) 
which is parallel to the axis and perpendicular to the 
force-plane, and along which the elements of the solid ex- 



FLEXURE. 



245 



perience no tension nor compression in an axial direction, 
this being called the Neutral Surface; 

(4.) The projection of the neutral surface upon the force 
plane (or a || plane) being called the Neutral Line or Elastic 
Curve, the bending or flexure of the piece is so slight that 
an elementary division, ds, of the neutral line may be put 
=dx, its projection on a line parallel to the direction of 
the axis before flexure ; 

(5.) The elements of the body contained between any 
two consecutive cross-sections, whose intersections with 
the neutral surface are the respective Neutral Axes of the 
sections, experience elongations (or contractions, accord- 
ing as they are situated on one side or the other of the 
neutral surface), in an axial direction, whose amounts are 
proportional to their distances from the neutral axis, and 
indicate corresponding tensile or compressive stresses ; 

(6.) F t =E c ; 

(7.) The dimensions of the cross-section are small com- 
pared with the length of the piece ; 

(8.) There is no shear perpendicular to the force plane 
on internal surfaces perpendicular to that plane. 

In the locality where any one of the external forces is 
applied, local stresses are of course induced which demand 
separate treatment. These are not considered at present. 

225. Illustration. — Consider the case of flexure shown in 
Fig. 221. The external forces are three (neglecting the 




Fig. 221. 



246 



MECHANICS OF ENGINEERING. 



weight of the beam), viz.: P lt P 2 , and P 3 . P x and P 3 are 
loads, P 2 the reaction of the support. 

The force plane is vertical. N Y L is the neutral line or 
elastic curve. NA is the neutral axis of the cross-section 
at m ; this cross-section, originally perpendicular to the 
sides of the prism, is during flexure "| to their tangent 
planes drawn at the intersection lines ; in other words, the 
side view QNB, of any cross-section is perpendicular to 
the neutral line. In considering the whole prism free we 
have the system P l9 P 2 , and P 3 in equilibrium, whence 
from JET==0 we have P 2 =Pi-hPs, and from JF(mom. about 
O)=0, PJ,3=Ph. Hence given Pj we may determine the 
other two external forces. A reaction such as P 2 is some- 
times called a supporting force. The elements above the 
neutral surface NiOLSaxs in tension ; those below in com- 
pression (in an axial direction). 

226. The Elastic Forces. — Conceive the beam in Fig. 221 
separated into two parts by any transverse section such 
as QA, and the portion ^ON, considered as a free body 
in Fig. 222. Of this free body the surface QAB is one of 



<Qdz 




;/ 



!/ 



F*G. 222. 



FLEXURE. 247 

the bounding surfaces, but was originally an internal sur- 
face of the beam in Fig. 221. Hence in Fig. 222 we must 
put in the stresses acting on all the dF'a or elements of area 
of QAB. These stresses represent the actions of the body 
taken away upon the body which is left, and according to 
assumptions (5), (6) and (8) consist of normal stresses (ten- 
sion or compression) proportional per unit of area, to the 
distance, z, of the dF's from the neutral axis, and of shear- 
ing stresses parallel to the force-plane (which in most 
cases will be vertical). 

The intensity of this shearing stress on any dF varies 
with the position of the dF with respect to the neutral 
axis, but the law of its variation will be investigated later 
■(§§ 253 and 254). These stresses, called the Elastic Forces 
of the cross-section exposed, and the external forces P x and 
P 2i form a system in equilibrium. We may therefore ap- 
ply any of the conditions of equilibrium proved in § 38. 



227. The Neutral Axis Contains the Centre of Gravity of the 
Cross-Section. — Fig. 222. Let e= the distance of the outer- 
most element of the cross-section from the neutral axis, and 
the normal stress per unit of area upon it be =p, whether 
tension or compression. Then by assumptions (5) and (6), 
§ 224, the intensity of normal stress on any dF is = -1 p 
and the actual 

normal stress on any dFis= — pdF . (1) 

This equation is true for dF's having negative sj's, i.e. 
on the other side of the neutral axis, the negative value 
of the force indicating normal stress of the opposite char- 
acter ; for if the relative elongation (or contraction) of two 
axial fibres is the same for equal z's, one above, the other 
below, the neutral surface, the stresses producing the 
changes in length are also the same, provided F t =F c ; see §§ 
184 and 201. 



248 MECHANICS OF ENGINEERING. 

For this free body in equilibrium put IX=0 (X is a 
horizontal axis). Put the normal stresses equal to their 
X components, the flexure being so slight, and the X com- 
ponent of the shears = for the same reason. This gives 
(see eq. (1) ) 

P± pdF= ; i.e. JL CdFz= ; or, t Fz=0 (2) 

In which z= distance of the centre of gravity of the cross - 
section from the neutral axis, from which, though un- 
known in position, the z's have been measured (see eq. 
(4) § 23). 

In eq. (2) neither p-^-e nor F can be zero .*. z must = ; 
i.e. the neutral axis contains the centre of gravity. Q. E. D. 
[If the external forces were not all perpendicular to the 
beam this result would not be obtained, necessarily.] 

228. The Shear. — The " total shear," or simply the 
" shear," in the cross-section is the sum of th.e vertical 
shearing stresses on the respective dF's. Call this sum 
J, and we shall have from the free body in Fig. 222, by 
putting 2Y=0 (Y being vertical) 

P 2 —JP l —J=0.\J=P 2 —P 1 . . (3) 

That is, the shear equals the algebraic sum of the ex- 
ternal forces acting on one side (only) of the section con- 
sidered. This result implies nothing concerning its mode 
of distribution over the section. 

229. The Moment. — By the "Moment of Flexure" or 
simply the Moment, at any cross- section is meant the sum 
of the moments of the elastic forces of the section, taking 
the neutral axis as an axis of moments. In this summa- 
tion the normal stresses appear alone, the shear taking no 
part, having no lever arm about the neutral axis. Hence, 
Fig. 222, the moment of flexure 



FLEXURE. 240 

This function, CdFz 2 , of the cross-section or plane figure 

is the quantity called Moment of Inertia of a plane figure, 
§ 85. For the free body in Fig. 222, by putting ^(mom.s 
about the neutral axis JSTA)=0, we have then 



±— — P l x l -\-P 2 x 2 =0, or in general, ±_ =31 . (5) 



in which M signifies the sum of moments, about the neutral 
axis of the section, of all the forces acting on the free body 
considered, exclusive of the elastic forces of the exposed 
section itself. 



230. Strength in Flexure. — Eq. (5) is available for solving 
problems involving the Strength of beams and girders, since 
it contains p, the greatest normal stress per unit of area to 
be found in the section. 

In the cases of the present chapter, where all the exter- 
nal forces are perpendicular to the prism or beam, and 
have therefore no components parallel to the beam, i.e. to 
the axis X, it is evident that the normal stresses in any 
section, as QB Fig. 222, are equivalent to a couple ; for the 
condition IX=0 falls entirely upon them and cannot be 
true unless the resultant of the tensions is equal, parallel, 
and opposite to that of the compressions. These two equal 
and parallel resultants, not being in the same line, form a 
couple (§ 28), which we may call the stress-couple. The 
moment of this couple is the " moment of flexure " p ~ , and 
it is further evident that the remaining forces in Fig. 222, 
viz.: the shear J and the external forces P L and P 2 , are 
equivalent to a couple of equal and opposite moment to 
the one formed by the normal stresses. 



250 MECHANICS OF ENGINEERING. 

231, Flexural Stiffness. — The neutral line, or elastic curve, 
containing the centres of gravity of all the sections, was 
originally straight ; its radius of curvature at any point, 
as N t Fig. 222, during flexure may be introduced as fol- 
lows. QB and U'V are two consecutive cross-sections, 
originally parallel, but now inclined so that the intersec- 
tion (7, found by prolonging them sufficiently, is the centre 
of curvature of the ds (put =dx) which separates them at 
N, and GG=p= the radius of curvature of the elastic 
curve at N, From the similar triangles U'UG&nd GNCwe 
have dl:dx : :e:p t in which dX is the elongation, U' U, of a 
portion, originally =dx, of the outer fibre. But the rela- 
tive elongation e= -j-~ of the latter is, by §184, within the 
ctx 

elastic limit, =^L.\ ~ —— and eq. (5) becomes 
E E p 

M=M .... (6) 



From (6) the radius of curvature can be computed. E= 
the value of E L =E C , as ascertained from experiments in 
bending. 

To obtain a differential equation of the elastic curve, (6) 
may be transformed thus, Fig. 223. The curve being very 

AXIS x flat, consider two consecutive 

ds's with equal dx's ; they may 
\ y be put = their dx's. Produce 

P^--^ — , the first to intersect the dy of the 

£§^^-?__fi&L__ second, thus cutting off the d 2 y, 
<S§J$$§> tfi7//A i.e. the difference between two 
^Sn^S* 90 q ^f//fy consecutive dy'a. Drawing a per- 
^ f * pendicular to each ds at its left 

extremity, the centre of curva- 
4p>^ ture G is determined by their in- 

£y tersection, and thus the radius 

£ of curvature p. The two shaded 

fig. 223. triangles have their small angles 



FLEXURE. 251 

«qual, and d?y is nearly perpendicular to the prolonged 
ds ; hence, considering them similar, we have 

p:dx::dx: d 2 y .*. — = -=£ 
and hence from eq. (6) we have 

(approx.) ±EI^=M . . (7) 

as a differential equation of the elastic curve. From this 
the equation of' the elastic curve may be found, the de- 
flections at different points computed, and an idea thus 
formed of the stiffness. All beams in the present chap- 
ter being prismatic and homogeneous both E and / are the 
same (i.e. constant) at all points of the elastic curve. In 
using (7) the axis Xmust be taken parallel to the length 
of the beam before flexure, which must be slight ; the 
minus sign in (7) provides for the case when a^y-^dx 2 is es- 
sentially negative. 

232. Resilience of Flexure. — If the external forces are made 
to increase gradually from zero up to certain maximum 
values, some of them may do work, by reason of their 
points of application moving through certain distances 
due to the yielding, or flexure, of the body. If at the be- 
ginning and also at the end of this operation the body is 
at rest, this work has been expended on the elastic resis- 
tance of the body, and an equal amount, called the work 
of resilience (or springing-back), will be restored by the 
elasticity of the body, if released from the external forces, 
provided the elastic limit has not been passed. The energy 
thus temporarily stored is of the potential kind ; see §§ 
148, 180, 196 and 218. 

232a. Distinction Between Simple, and Continuous, Beams (or 
"Girders"). — The external forces acting on a beam consist 



252 



MECHANICS OF ENGINEERING. 



generally of the loads and the " reactions " of the sup- 
ports. If the beam is horizontal and rests on two supports, 
only, the reactions of those supports are easily found by 
elementary statics [§ 36] alone, without calling into ac- 
count the theory of flexure, and the beam is said to be a 
Simple Beam, or girder ; whereas if it is in contact with 
more than two supports, being " continuous," therefore, 
over some of them, it is a Continuous Girder (§271). The 
remainder of this chapter will deal only with simple 
beams. 



ELASTIC CURVES. 

233. Case I. Horizontal Prismatic Beam, ^Supported at Both 
Ends, With a Central Load, Weight of Beam Neglected. — Fig. 
224. First considering the whole beam free, we find each 




^ rr:_.if 



Fig. 224. § 233. 

reaction to be =%P. AOB is the neutral line ; required 
the equation of the portion OB referred to as an origin, 
and to the tangent line through as the axis of X To 
do this consider as free the portion mB between any sec- 
tion m on the right of and the near support, in Fig. 
225. The forces holding this free body in equilibrium 



BAat 




Fig. 225. 



Fig. 226. 



ELASTIC CURVES. 253 

are the one external force y£P y and the elastic forces act- 
ing on the exposed surface. The latter consist of J t the 
shear, and the tensions and compressions represented in 
the figure by their equivalent " stress-couple." Selecting 
N, the neutral axis of ra, as an axis of moments (that J 
may not appear in the moment equation) and putting 
2 (mom) =0 we have 

- 2 \ 2 ) dx 2 dx 2 2 V 2 ) 



Fig. 226 shows the elastic curve OB in its purely geomet- 
rical aspect, much exaggerated. For axes and origin as in 
figure d 2 y-^dx 2 is positive. 

Eq. (1) gives the second ^-derivative of y equal to a 
function of x. Hence the first cc-derivative of y will be 
equal to the cc-anti-derivative of that function, plus a con- 
stant, C. (By anti-derivative is meant the converse of de- 
rivative, sometimes called integral though not in the sense 
of summation). Hence from £L) we have {EI being a con- 
stant factor remaining undisturbed) 



JldjL=*L(l.x — t)+C . . (2)' 



dx 2 V2 2 



(2)' is an equation between two variables dy-^-dx and x, and 
holds good for any point between and B ; dy-^-dx de- 
noting the tang, of «, the slope, or angle between the tan- 
gent line and X. At the slope is zero, and x also zero ; 
hence at (2)' becomes 

EIxO=0—0+C 

which enables us to determine the constant C, whose value 
must be the same at as for all points of the curve. 
Hence (7=0 and (2)' becomes 



254 MECHANICS OF ENGINEERING. 

(4-H?) • • • (2 > 



dx 2 



from whicli the slope, tan. «, (or simply «, in tt -measure ; 
since the angle is small) may be found at any point. Thus 
at B we have x=ffl and dy-r-dx=a lt and 

. _ 1 PI 2 

•• ttl "l6 ' EI 

Again, taking the #-anti -derivative of both members of eq. 
(2) we have 

EIy= i (¥~?) +<7 ' • • • (3y 

and since at both x and y are zero, C is zero. Hence 
the equation of the elastic curve OB is 

**-f &-t) ' • ' (3) 

To compute the deflection of from the right line join- 
ing A and B in Fig. 224, i.e. BK, =d, we put x—y 2 l in (3), y 
being then =d, and obtain 

**=*=&' is ■ • • < 4 > 

Eq. (3) does not admit of negative values for x ; for if 
the free body of Fig. 225 extended to the left of 0, the ex- 
ternal forces acting would be P, downward, at ; and %P, 
upward, at B, instead of the latter alone ; thus altering 
the form of eq. (1). From symmetry, however, we know 
that the curve AO, Fig. 224, is symmetrical with OB about 
the vertical through 0. 



ELASTIC CURVES. 255 

233a. Load Suddenly Applied. — Eq. (4) gives the deflection 
d corresponding to the force or pressure P applied at the 
middle of the beam, and is seen to be proportional to it. 
If a load G hangs at rest from the middle of the beam, 
P= Q ; but if the load G, being initially placed at rest 
upon the unbent beam, is suddenly released from the ex- 
ternal constraint necessary to hold it there, it sinks and 
deflects the beam, the pressure P actually felt by the beam 
varying with the deflection as the load sinks. What is 
the ultimate deflection d m ? Let P m = the pressure be- 
tween the load and the beam at the instant of maximum 
deflection. The work so far done in bending the beam 
= }£P m d m . The potential energy given up by the load 
= Gd mi while £he initial and final kinetic energies are both 
nothing. 

.-. Gd m =y 2m Pd m . . (5) 

That is, P m =26r. Since at this instant the load is sub- 
jected to an upward force of 2G and to a downward force 
of only G (gravity) it immediately begins an upward mo- 
tion, reaching the point whence the motion began, and 
thus the oscillation continues. "We here suppose the elas- 
ticity of the beam unimpaired. This is called the " sud- 
den " application of a load, and produces, as shown above, 
double the pressure on the beam which it does when grad- 
ually applied, and a double deflection. The work done 
by the beam in raising the weight again is called its re- 
silience. 

Similarly, if the weight G is allowed to fall on the mid- 
dle of the beam from a height h, we shall have 

Gx(h+d m ), or approx., Gh=)4P m d m ; 

and hence, since (4) gives d m in terms of P m , 




256 MECHANICS OF ENGINEERING. 

This theory supposes the mass of the beam small com- 
pared with the falling weight. 

234. Case II. Horizontal Prismatic Beam, Supported at Both 
Ends, Bearing a Single Eccentric Load. Weight of Beam Neg- 
lected. — Fig. 227. The reactions 
of the points of support, P Q and 
Pu are easily found by consider- 
ing the whole beam free, and put- 
ting first Ifmom.),, =0, whence P x 
~' It I =Pl-t-l L , and then ^(mom.^^O, 

no 227. whence P =P(l l —l)+l 1 . P and 

Pi will now be treated as known quantities. 

The elastic curves OG and CB, though having a common 
tangent line at G (and hence the same slope « c ), and a com- 
mon ordinate at (7, have separate equations and are both 
referred to the same origin and axes, as shown in the 
figure. The slope at 0, «o, and that at B,a it are unknown 
constants, to be determined in the progress of the work. 

Equation of OC. — Considering as free a portion of the 
beam extending from B to a section made anywhere on 
OG, x and y being the co-ordinates of the neutral axis of 
that section, we conceive the elastic forces put in on the 
exposed surface, as in the preceding problem, and put 
2'(mom. about neutral axis of the section) =0 which gives 
(remembering that here d 2 y-±-dx 2 is negative.) 

EI % =P(l-x)-P 1 (l l -x) ; . . (1) 
a x 

whence, by taking the x anti-derivatives of both members 

M i =r(i*~)-m*- -£-)+ o 

To find G, write out this equation for the point 0, where 
dy^-dx=a and x=0, and we have G—EIa^\ hence the 
equation for slope is 



FLEXURE ELASTIC CUBVES. 257 

EI^=P(lx--^)-P l (l 1 x-^)+EIa . (2) 

Again taking the x anti-derivatives, we have from (2) 

Ely =P gL- ^y-pM~ £ Y ma &+(C'=0) (3) 

(at Oboth x and 2/ are =0 .'. (7'=0). In equations (1), (2), 
and (3) no value of x is to be used <0 or >?, since for 
points in CB different relations apply, thus 

Equation of CB. — Fig. 227. Let the free body extend 
from B to a section made anywhere on CTJfmoms.), as 
before, =0, gives 

m % =- p ^- x > ■ • - w 

(N.B. In (4), as in (1), EWy-h-dx 2 is written equal to a neg- 
ative quantity because itself essentially negative ; for the 
curve is concave to the axis X in the first quadrant of the 
co-ordinate axes.) 

From (4) we have in the ordinary way (x-anti-deriv.) 

EI^-P^x -^)+C" . . (5/ 

To determine (7", consider that the curves CB and OC 
have the same slope (dy-r-dx) at C where x=l; hence put 
x=l in the right-hand members of (2) and of (5)' and 
equate the results. This gives C r " = J^PZ 2 +^/a an( ^ •'• 

ei % = t + M *>- p iliX ~P • (5) 



and .: Ely =^x+E1o V jc-Ps£-~1 +0'" . (6)' 

A A D 



258 MECHANICS OF ENGINEERING. 

At (7, where x=l, both, curves have the same ordinate^ 
hence, by putting x=l in the right members of (3) and (6)' 
and equating results, we obtain C'"= — %Pl*. .*. (6)' be 
comes 

Ely^PVx+EIa&^-P, \^-^~^ . (6) 

as the Equation of CB, Fig. 227. But Oq is still an unknown 
constant, to find which write out (6) for the point B where 
x=l, and y=0j whence we obtain 

^ = eil [PP_3jPKl+2PA3] * * (7) 

« x = a similar form, putting P for P ly and (li — I) for I. 

235. Maximum Deflection in Case II — Fig. 227. The or- 
dinate y m of the lowest point is thus found. Assuming 
l>}&\> it will occur in the curve OC. Hence put the* 
dy-h-dx of that curve, as expressed in equation (2), =0. 
Also for « write its value from (7), having iputP 1 =Pl-±-l li 
and we have 

P(lx-~)-P± (hx— *)+ y 6 ^(F-31l l +2in-0 

whence [a? for max. y]=^/y 1(2^—1) 

Now substitute this value of x in (6), also «q from (7), and 
putPi=P^?i, whence 

p 

Max. Deflec.=2/max=7 9 • ^T C^ 3 — 3Z 2 Z 1 +2Z? 1 2 ] ^/y^l.—l). 

236. Case III. Horizontal Prismatic Beam Supported at Both 
Ends and Bearing a Uniformly Distributed Load along its Whole 
Length. — (The weight of the beam itself, if considered, 



FLEXURE. ELASTIC CURVES. 



259 



constitutes a load of this nature.) Let 1= the length 
of the beam and iv= the weight, per unit of length, 
of the loading ; then the load coming upon any length x 
will be =wx, and the whole load =id. By hypothesis w 
is constant. Fig. 228. From symmetry we know that the 



W«M>? 



ux dtpttu I 




Qj=fe2=: 




Fig. 228. 

reactions at A and B are each =}4wl, that the middle of 
the neutral line is its lowest point, and the tangent line at 
is horizontal. Conceiving a section made at any point 
m of the neutral line at a distance x from 0, consider as 
free the portion of beam on the right of m. The forces 
holding this portion in equilibrium are y^wl, the reaction 
at B ; the elastic forces of the exposed surface at m, viz.: 
the tensions and compressions, forming a couple, and J 
the total sheor ; and a portion of the load, iv( l / 2 l — x). The 
sum of the moments of these latter forces about the neu- 
tral axis of m, is the same as that of their resultant ; (i.e., 
their sum, since they are parallel), and this resultant acts in 
the middle of the length y 2 l — x. Hence the sum of these 
moments =w(*4l — x))4{yi — x). Now putting I (mom. 
about neutral axis of m) =0 for this free body, we have 



«& 



4wi( y 2 1— x)— y 2 w( y 2 1— xj 



i.e. 



ax 2 



wiyp—x 2 ) 



(i) 



260 MECHANICS OF ENGINEERING. 

Taking the a>anti-derivative of both sides of (1), 

EI Tx = I /M I A^- 1 A^)+(C=0) (2) 

as the equation of slope. (The constant is =0 since at 
both dy-i-dx and x are =0.) From (2), 

^=|(>^ 2 -7i2* 4 )+[C'=0] . . (3) 

which is the equation of the elastic curve ; throughout, 
i.e., it admits any yalue of x from x=-\-y^l to x= — y 2 l. 
This is an equation of the fourth degree, one degree high- 
er than those for the Curves of Cases I and II, where 
there were no distributed loads. If w were not constant, 
but proportional to the ordinates of an inclined right line, 
eq. (3) would be of the fifth degree; if w were propor- 
tional to the vertical ordinates of a parabola with axis 
vertical, (3) would be of the sixth degree ; and so on. 

By putting x=*4l in (3) we have the deflection of be- 
low the horizontal through A and B, viz.: (with W= total 
load =wl) 

384 ' EI 384 # EI { ' 

237. Case IV. Cantilevers. — A horizontal beam whose only 
support consists in one end being built in a wall, as in 
Fig. 229(a), or supported as in Fig. 
229(&) is sometimes called a canti- 
lever. Let the student prove that in 
Fig. 229(a) with a single end load P, 
the deflection of B below the tangent 
at Ois d=j4 ) Pl 3 -—EI;the same state- 
ment applies to Fig. 229(6), but the 
tangent at is not horizontal if the 
beam was originally so. It can also 
fig. 229. be p roV ed that the slope at B, Fig. 

229(a) (from the tangent at 0) is 




FLEXUBE ELASTIC CURVES. 261 

PI 2 

The greatest deflection of the elastic curve from the right 
line joining AB, in Fig. 229(6), is evidently given by the 
equation for y max. in § 235, by writing, instead of P of 
that equation, the reaction at in Fig. 229(6). This assumes 
that the max. deflection occurs between A and 0. If it 
occurs between and B put (li—l) for I. 

If in Fig. 229(a) the loading is uniformly distributed 
along the beam at the rate of w pounds per linear unit, 
the student may also prove that the deflection of B below 
the tangent at is 

238. Case V. Horizontal Prismatic Beam Bearing Equal Ter- 
minal Loads and Supported Symmetrically at Two Points.—- 
Fig. 231. "Weight of beam neglected. In the preceding 
cases we have made use of the approximate form EId?y-h-dx 2 
in determining the forms of elastic curves. In the present 



r^~T=] 



pi 

1- 

Fig. 231. 




case the elastic curve from to C is more directly dealt 
with by employing the more exact expression El-r-p (see 
§ 231) for the moment of the stress-couple in any section. 
The reactions at and G are each = P, from symmetry. 
Considering free a portion of the beam extending from A 
to any section m between and C (Fig. 232) we have, by 
putting 2 (mom. about neutral axis of m)=0, 



262 MECHANICS OF ENGINEEKING. 

That is, the radius of curvature is the same at all points 
of 0(7; in other words OC is the arc of a circle with the 
above radius. The upward deflection of F from the right 
line joining and G can easily be computed from a knowl- 
edge of this fact. This is left to the student as also the 
value of the slope of the tangent line at (and C). The 
deflection of D from the tangent at C= 1 / s PI b -t-UI j as ip 
Fig. 229(a). 



SAFE LOADS IN FLEXURE. 

239. Maximum Moment. — As we examine the different sec- 
tions of a given beam under a given loading we find differ- 
ent values of p, the normal stress per unit of area in the 
outer element, as obtained from eq. (5) § 229, viz.: 

pL=m. . . . . (i) 

in which 7" is the " Moment of Inertia " (§ 85) of the plane 
figure formed by the section, about its neutral axis, e the 
distance of the most distant (or outer) fibre from the neu- 
tral axis, and M the sum of the moments, about this neu- 
tral axis, of all the forces acting on the free body of which 
the section in question is one end, exclusive of the stresses 
on the exposed surface of that section. In other words 
M is the sum of the moments of the forces which balance 
the -stresses of the section, these moments being taken 
about the neutral axis of the section under examination. 
For the prismatic beams of this chapter e and /are the 
.same at all sections, hence p varies with M and becomes a 
maximum when M is a maximum. In any given case the 
location of the "dangerous section" or section of maximum 
M, and the amount of that maximum value may be deter- 
mined by inspection and trial, this being the only method 
(except by graphics) if the external forces are detached. 



FLEXURE SAFE LOADS. 263 

If, however, the loading is continuous according to a de- 
finite algebraic law the calculus may often be applied, 
taking care to treat separately each portion of the beam 
between two consecutive reactions of supports, or detached 
loads. 

As a graphical representation of the values of M along 
the beam in any given case, these values may be conceived 
laid off as vertical ordinates (according to some definite 
scale, e.g. so many inch-lbs. of moment to the linear inch 
of paper) from a horizontal axis just below the beam. If 
the upper fibres are in compression in any portion of the 
beam, so that that portion is convex downwards, these or- 
dinates will be laid off below the axis, and vice versa ; for 
it is evident that at a section where M=Q, p also =0, i.e., 
the character of the normal stress in the outermost fibre 
changes (from tension to compression, or vice versa) when 
M changes sign. It is also evident from eq. (6) § 231 that 
the radius of curvature changes sign, and consequently the 
curvature is reversed, when If changes sign. These mo- 
ment ordinates form a Moment Diagram, and the extremities 
a Moment Curve. 

The maximum moment, 3f m , being found, in terms of 
the loads and reactions, we must make the p of the " dan- 
gerous section," where M= M m , equal to a safe value R' 9 
and thus may write 

— =M m . . . . (2) 



Eq. (2) is available for finding any one unknown quanti- 
ty, whether it be a load, span, or some one dimension of 
the beam, and is concerned only with the Strength, and not 
with the stiffness of the beam. If it is satisfied in any 
given case, the normal stress on all elements in all sections 
is known to be = or <i?', and the design is therefore safe 
in that one respect. 

As to danger arising from the shearing stresses in any 



264 



MECHANICS OF ENGINEERING. 



section, the consideration of the latter will be taken up in 
a subsequent chapter and will be found to be necessary 
only in beams composed of a thin web uniting two flanges. 
The total shear, however, denoted by J, bears to the mo- 
ment M, an important relation of great service in deter- 
mining M m . This relation, therefore, is presented in the 
next article. 

240. The Shear is the First x-Derivative of the Moment. — 
Fig. 233. (x is the distance of any section, measured parallel 
wdx \/ to the beam from an arbitrary 

pd'p origin). Consider as free a ver- 
tical slice of the beam included 
between any two consecutive 
vertical sections whose distance 
apart is dx. The forces acting 
are the elastic forces of the two 
internal surfaces now laid bare, 
and, possibly, a portion, wdx, 
of the loading, which at this 
part of the beam has some intensity =w lbs. per running 
linear unit. Putting ^(mom. about axis .iV')=0we have 
(noting that since the tensions and compressions of section 
iV" form a couple, the sum of their moments about N' is 
just the same as about JV,) 



pdF 



-pd? 




£l-J^ + Jdx+wdx.^-=0 



But Pl=M, the Moment of the left hand section,^- =ifcp, 
e e 

that of the right ; whence we may write, after dividing 

through by dx and transposing, 

dM 



M'—M T ^ dx 

-d^ =J+w T 



i.e.,- 



dx 



J; 



(3). 



for to % vanishes when added to the finite J, and M' — M= 
dM= increment of the moment corresponding to the incre- 
ment, dx, of x. This proves the theorem. 



FLEXURE. SAFE LOADS. 2G5 

Now the value of x which renders M a maximum or 
minimum would be obtained by putting the derivative 
dM-h-dx= zero ; hence we may state as a 

Corollary. —At sections where the moment is a maximum 
or minimum the shear is zero. 

The shear J at any section is easily determined by con- 
sidering free the portion of beam from the section to either 
end of the beam and putting ^(vertical components)=0. 

In this article the words maximum and minimum are 
used in the same sense as in calculus ; i.e., graphically, 
they are the ordinates of the moment curve at points 
where the tangent line is horizontal. If the moment curve be 
reduced to a straight line, or a series of straight lines, it 
has no maximum or minimum in the strict sense just 
stated ; nevertheless th$ relation is still practically borne 
out by the fact that at the sections of greatest and least 
ordinates in the moment diagram the shear changes sign 
suddenly. This is best shown by drawing a shear diagram, 
whose ordinates are laid off vertically from a horizontal 
axis and under the respective sections of the beam. They 
will be laid off upward or downward according as J is 
found to be upward or downward, when the free body con- 
sidered extends from the section toward the right. 

In these diagrams the moment ordinates are set off on 
an arbitrary scale of so many inch-pounds, or foot-pounds, 
to the linear inch of paper ; the shears being simply 
pounds, or some other unit of force, on a scale of so many 
pounds to the inch of paper. The scale on which the 
beam is drawn is so many feet, or inches, to the inch of 
paper. 

241. Safe Load at the Middle of a Prismatic Beam Support- 
ed at the Ends. — Fig. 234. The reaction at each support 
is y^P. Make a section n at any distance x<~L from B* 
Consider the portion nB free, putting in the proper elas- 
tic and external forces. The weight of beam is neglected. 
From JT(mom. about n)=Q we have 



266 



MECHANICS OF ENGINEERING. 



e 2 

Evidently ilt^is proportional to x, and the ordinates repre- 
senting it will therefore be limited by the straight line 




,Fig. 234. 



B'R, forming a triangle B'RA'. From symmetry, another 
triangle O'RA' forms the other half of the moment dia- 
gram. From inspection, the maximum M is seen to be in 
the middle where x= yil, and hence 



(Mm&TL.)=M m =%Pl 



(1) 



Again by putting 2Yvert. compons.)=0, for the free body 
nB we have 

*5 



and must point downward since £ points upward. Hence 
the shear is constant and = ^Pat any section in the right 
hand half. If n be taken in the left half we would have, 
nB being free, from Invert. com.)=0, 

j=p—y 2 p=y 2 p 



FLEXURE. SAFE LOADS. 267 

the same numerical value as before ; but J" must point up- 
ward, since t at B and J at n must balance the downward 
P at A. At A, then, the shear changes sign suddenly, 
that is, passes through the value zero ; also at A, M is a 
maximum, thus illustrating the statement in § 240. Notice 
the shear diagram in Fig. 234. 

To find the safe load in this case we write the maximum 
value of the normal stress, p,= R\ a safe value, (see table 
in a subsequent article) and solve the equation for P. 
But the maximum value of p is in the outer fibre at A> 
since M for that section is a maximum. Hence 



— = %n (2) 



is the equation for safe loading in this case, so far as the 
normal stresses in any section are concerned. 

Example. — If the beam is of wood and has a rectangu- 
lar section with width b= 2 in., height h*= 4 in., while its 
length 1= 10 ft., required the safe load, if the greatest nor- 
mal stress is limited to 1,000 lbs. per sq. in. Use the 
pound and inch. From § 90 1=% bh*=%X 2x64= 10.66 
biquad. inches, while e=£.= 2 in. 



,. P= ^4x1,000x10.66 =177 , 7 lb , 
le 120x2 



242. Safe Load Uniformly Distributed along a Prismatic Beam 
Supported at the Ends.— Let the load per lineal unit of the 
length of beam be =w (this can be made to include the 
weight of the beam itself). Fig. 235. From symmetry, 

each reaction = y 2 wl. For the free body nO we have, put- 
ting J(mom. about n)=0, : : 



2(58 



MECHANICS OF ENGINEERING. 



which gives Mlor any section by making x vary from 
to I. Notice that in this case the law of loading is con- 
tinuous along the whole length, and that hence the mo- 
ment curve is continuous for the whole length. 




Fig. 235. 



To find the shear J, at n, we may either put J(vert. com- 
pons.)=0 for the free body, whence J= y 2 wl — tox, and must 
therefore be downward for a small value of x ; or, employ- 
ing § 240, we may write out dM~dx, which gives 



"-*►-•■» 



a: 



the same as before. To find the max. 31, or 3I m , put «7=0, 
which gives x=yi. This indicates a maximum, for when 
substituted in d z M-h-dx 2 ^ i.e., in — to, a negative result is 
obtained. Hence M m occurs at the middle of the beam and 
its value is 



M m = y 8 wl 2 ; 



EI 



= y sW l2 =1/ £ W i 



(2) 



the equation of safe loading. W= total lo'a,d=wl. 

It can easily be shown that the moment curve is a por- 



FLEXURE. SAFE LOADS. 



269 



tion of a parabola, whose vertex is at A" under the mid- 
dle of the beam, and axis vertical. The shear diagram 
consists of ordinates to a single straight line inclined to 
its axis and crossing it, i.e., giving a zero shear, under the 
middle of the beam, where we find the max. 31. 

If a frictionless dove-tail joint with vertical faces were 
introduced at any locality in the beam and thus divided 
the beam into two parts, the presence of J would be made 
manifest by the downward slipping of the left hand part 
on the right hand part if the joint were on the right of the 
middle, and vice versa if it were on the left of the middle. 
This shows why the ordinates in the two halves of the 
shear diagram have opposite signs. The greatest shear 
is close to either support and is J m =^}^icl. 

243. Prismatic Beam Supported at its Extremities and Loaded 
in any Manner. Equation for Safe Loading. — Fig. 236. Given 

the loads P lt P 2 , and P 3 , whose 
distances from the right sup- 
^h Fi EW-J-H P or * are t ha and l 3 ; ,required 

— p the equation for safe loading ; 
i.e., find 3I m and write it = 
R'l+e. 

If the moment curve were 
continuous, i.e., if M were a 
continuous function of x from 
end to end of the beam, we 
could easily find M m by making 
dJl- : rdx=0, i.e., J=0, and sub- 
stitute the resulting value of x in the expression for 31. 
But in the present case of detached loads, J" is not zero, 
necessarily, at any section of the beam. Still there is 
some one section where it changes sign, i.e., passes sud- 
denly through the value zero, and this will be the section 
of greatest moment (though not a maximum in the strict 
sense used in calculus). By considering any portion n O 
as free, J" is found equal to the Reaction at Diminished by 
the Loads Occuring "Between n and 0. The reaction at B is 



r r r 




Fig. 236. 



270 MECHANICS OF ENGINEERING. 

obtained by treating the whole beam as free (in which case 
no elastic forces come into play) and putting ^(mom. 
about O)=0; while that at 0,=P Q =P 1 +P 2 +P ir -P B 
If n is taken anywhere between and E, J=P 

E " F f J=P -P l 
F " H,J=P -P 1 -P 2 
H " B, J=P -P 1 -P 2 -P 3 
This last value of J" also = the reaction at the other 
support, B. Accordingly, the shear diagram is seen to 
consist of a number of horizontal steps. The relation 
J—dM-r-dx is such that the slope of the moment curve is 
proportional to the ordinate of the shear diagram, and 
that for a sudden change in the slope of the moment curve 
there is a sudden change in the shear ordinate. Hence in 
the present instance, J being constant between any two 
consecutive loads, the moment curve reduces to a straight 
line between the same loads, this line having a different 
inclination under each of the portions into which the beam 
is divided by the loads. Under each load the slope of the 
moment curve and the ordinate of the shear diagram change 
suddenly. In Fig. 236 the shear passes through the value 
zero, i.e., changes sign, at F; or algebraically we are sup- 
posed to find that P —P 1 is + while P Q —P l —P 2 is — , in 
the present case. Considering F0 } then, as free, we find 
M m to be 

M m =PJj 2 —Pi(l 2 —l\) and the equation for safe loading is 

—-PA-pfa-td (i) 

e 

(i.e., if the max. Mis at F). It is also evident that the 
greatest shear is equal to the reaction at one or the other 
support, whichever is the greater, and that the moment 
at either support is zero. 

The student should not confuse the moment curve, which 



FLEXURE. SAFE LOADS. 



271 



is entirely imaginary, with the neutral line (or elastic 
curve) of the beam itself. The greatest moment is not 
necessarily at the section of maximum deflection of the 
neutral line (or elastic curve). 

For the case in Fig. 236 we may therefore state that the 
max. moment, and consequently the greatest tension or 
compression in the outer fibre, will be found in the sec- 
tion under that load for which the sum of the loads (in- 
cluding tins' load itself) between it and either support first 
equals or exceeds the reaction of that support. The 
amount of this moment is then obtained by treating as free 
either of the two portions of the beam into which this 
section divides the beam. 

244. Numerical Example of the Preceding Article. — Fig. 237. 
Given P l9 P 2 , P 3 , equal to j4 ton, 1 ton, and 4 tons, re- 



K 



i T0N p 



Li 



T 11 7~EJ 

|— 60 — ,±3 




spectively ; Z x =5 feet, ? 2 =7 feet, and 4=10 feet ; while the 
total length is 15 feet. The beam is of timber, of rectan- 
gular cross-section, the horizontal width being &=10 
inches, and the value of B! (greatest safe normal stress), 
= 1/2, ton per sq. inch, or 1,000 lbs. per sq inch. 



272 MECHANICS OF ENGINEERING. 

Required the proper depth h lor the beam, for safe load- 
ing. 

Solution. — Adopting a definite system of units, viz., the 
inch-ton-second system, we must reduce all distances such 
as I, etc., to inches, express all forces in tons, write R' = y 2 
(tons per sq. inch), and interpret all results by the same sys- 
tem. Moments will be in inch-tons, and shears in tons. 
[N. B. In problems involving the strength of materials 
the inch is more convenient as a linear unit than the foot, 
since any stress expressed in lbs., or tons, per sq. inch, is 
numerically 144 times as small as if referred to the square 
foot.] 

Making the whole beam free, we have from moms, about 
O, P B ~ [^x60+1x84+4x120]=3.3 tons.-. P =5.5— 
3.3=2.2 tons. 

The shear anywhere between and E is J— P =2.2 tons. 

E and F is J =2.2— y=l.l 

tons. 
The shear anywhere between F and His J =2.2 — y — 1 = 

0.7 tons. 
The shear anywhere between H and B is J = 2.2 — y 2 — 1 

—4 =—3.3 tons. 
Since the shear changes sign on passing H, .-. the max. 
moment is at H; whence making HO free, we have 
M at H=M m =2.2 x 120—^ x 60—1 x 36 =198 inch -tons. 

r>/ r 

F or safety M m must = - — , in which B ' = % ton per sq. 

e 

inch, e — y 2 \L — y 2 of unknown depth of beam, and I, §90, = 
I &A 3 , with &=10 inches 

12 /w h 



^X10A 3 =198; or ¥ =237.6.-. 7^=15.4 inches. 



245. Comparative Strength, of Rectangular Beams. — For such 
a beam, under a given loading, the equation for safe load- 
ing is 

?I=M m i. e. % B' b¥=M m .... (1) 



FLEXURE. SAFE LOADS. 273 

whence the following is evident, (since for the same length, 
mode of support, and distribution of load, M m is propor- 
tional to the safe loading.) 

For rectangular prismatic beams of the same length, 
same material, same mode of support and same arrange- 
ment of load : 

(1) The safe load is proportional to the width of beams 
having the same depth (h). 

(2) The safe load is proportional to the square of the 
depth of beams having the same width (b). 

(3) The safe load is proportional to the depth of beams 
having the same volume (i. e. the same bh). 

(It is understood that the sides of the section are hori- 
zontal and vertical respectively and that the material is 
homogeneous.) 

246. Comparative Stiffness of Rectangular Beams.-- Taking the 
deflection under the same loading as an inverse measure 
of the stiffness, and noting that in §§ 233, 235, and 236, 
this deflection is inversely proportional to I = ^ 2 bh? = 
the "moment of inertia" of the section about its neutral 
axis, we may state that : 

For rectangular prismatic beams of the same length, 
same material, same mode of support, and same loading : 

(1) The stiffness is proportional to the width for beams 
of the same depth. 

(2) The stiffness is proportional to the cube of the 
height for beams of the same width (b). 

(3) The stiffness is proportional to the square of the 
depth for beams of equal volume (bhl). 

(4) If the length alone vary, the stiffness is inversely 
proportional to the cube of the length. 

247. Table of Moments of Inertia. — These are here recapitu- 
lated for the simpler cases, and also the values of e. the 
distance of the outermost fibre from the axis. 

Since the stiffness varies as /(other things being equal), 



274 



MECHANICS OF ENGINEERING. 



while the strength varies as I-^-e, it is evident that a 
square beam has the same stiffness in any position (§89), 
while its strength is greatest with one side horizontal, for 
then e is smallest, being —y^b. 

Since for any cross-section I— CdF z 2 , in which z=the 

distance of any element, dF, of area from the neutral axis, 
a beam is made both stiffer and stronger by throwing 
most of its material into two flanges united by a vertical 
web, thus forming a so-called " I-beam " of an I shape. But 
not without limit, for the web must be thick enough to 
cause the flanges to act together as a solid of continuous 
substance, and, if too high, is liable to buckle sideways, 
thus requiring lateral stiffening. These points will be 
treated later. 



*,— 




t 

i 

i 




FFR 






r 












i 



(a) (b) 

Fia. 238. 




SECTION. 


I 


e 


Rectangle, width = b, depth = h (vertical) 


Via M» 


% h 


Hollow Rectangle, symmet. about neutral axis. See I 
Fig. 238 (a) j 


Via [&i hS-b 2 h\1 


XK 


Triangle, width —b, height = h, neutral axis parallel I 
to base (horizontal). f 


7 36 bh* 


%fo 


Circle of radius r 


%nr* 


r 


Ring of concentric circles. Fig. 238 (b) 


J AMr A !-r* 2 ) 


U 


Rhombus; Fig. 238 (c) h = diagonal which is vertical. 


748 bh* 


%h 


Square with side 6 vertical. 


Via b* 


y 2 b 


" " " b at 45° with horiz. 


Via b* 


HbV2 



248. Moment of Inertia of I-beams, Box-girders, Etc. — In 
common with other large companies, the N. J. Steel and 



FLEXURE. SAFE LOADS. 



275 



Iron Co. of Trenton, N. J. (Cooper, Hewitt & Co.) manu- 
facture prismatic rolled beams of wrought-iron variously 
called /-beams, deck -beams, rails, and " shape iron," (in- 
cluding channels, angles, tees, etc., according to the form 
of section.) See fig. 239 for these forms. The company 




T 



channel. deck-beam. rail. 
Fig. 239. 



ANGLE-IRON. 



publishes a pocket-book giving tables of quantities rela- 
ting to the strength and stiffness of beams, such as the 
safe loads for various spans, moments of inertia of their 
sections in various positions, etc., etc. The moments of 
inertia of /-beams and deck -beams are computed accord- 
ing to §§ 92 and 93, with the inch as linear unit. The 
/-beams range from 4 in. to 20 inches deep, the deck- 
beams being about 7 and 8 in. deep. 

For beams of still greater stiffness and strength com- 
binations of plates, channels, angles, etc., are riveted to- 
gether, forming " built-beams," or " plate girders." The 
proper design for the riveting of such beams will be ex- 
amined later. For the present the parts are assumed to 
act together as a continuous mass. For example, Fig. 240 
shows a " box-girder," formed of two " channels " and 
two plates riveted together. If the axis of symmetry, N, 

\r — —6 1 is to be horizontal it becomes the neu- 

Hff tral axis. Let C= the moment of iner- 
tia of one channel (as given in the 
pocket-book mentioned) about the axis 
N perpendicular to the web of the chan- 
nel. Then the total moment of inertia of 
the combination is (nearly) 



m 



m 



£i 



Fig. 240. 



z N =2c+2btd 2 — wt\d— y 2 ty 



(i) 



27G 



MECHANICS OP ENGINEERING. 



In (1), b, t, and d are the distances given in Fig. 240 (d ex- 
tends to the middle of plate) while d' and t' are the length 
and width of a rivet, the former from head to head 
{i.e., d' and t' are the dimensions of a rivet-hole). 

For example, a box-girder of wroraght-iron is formed of 
two 15-inch channels and two plates 10 inches wide and 1 
rivet holes ^ in. wide and 1}{ in. long. 
t=l: d=S; t' = 3/; and d '=13/ inches. 



inch thick, 
That is, b- 



the 
=10 



Also from the pocket-book we find that for the channel in 
question, (7=376 biquadratic inches. Hence, eq. (1) 

^ = 752+2x10x1x64— 4x1x^(8— ^) 2 =1737biquadr.in. 



Also, since in this instance e = 8)4 inches, and 12000 
lbs. per sq. inch (or 6 tons per sq. in.) is the value for B' 
>(=greatest safe normal stress on the outer element of any 
cross-section) used by the Trenton Co. (for wrought iron), 



we have 



BI 12000x1737 



8.5 



=2451700 inch-lbs. 



That is, the box-girder can safely bear a maximum mo- 
ment, M m , = 2451700 inch-lbs. = 1225.8 inch-tons, as far 
as the normal stresses in any section are concerned. 
(Proper provision for the shearing stresses in the section, 
and in the rivets, will be considered later). 



249. 



Strength of Cantilevers. — In Fig. 241 with a single 
i ,_( v\mw? concentrated load P at the 

i^j VJ 1 1 I I 1 j | projecting extremity, we 
f° easily find the moment at 
n to be M =Px, and the 
max. moment to occur at 
the section next the wall, 
p its value being M m =Pl. 
The shear, J, is constant, 
fig. 242. an( j = p a t all sections. 





Fig. 241. 



The moment and shear diagrams are drawn in accordance 
with these results. 



FLEXUEE. SAFE LOADS. 277 

If the load TV = id is uniformly distributed on the can- 
tilever, as in fig. 242, by making nO free we have, putting 
I(mom. about n) = 0, 

v —=wx . | rJf=y 2 tcx i .-. M m =y 2 id 2 = y 2 m. 

Hence the moment curve in a parabola, whose vertex is at 
Of and axis vertical. Putting 1 (vert, compons.) = we 
obtain J = tax. Hence the shear diagram in a triangle, 
and the max. J= id = W. 

250. Resume' of the Four Simple Cases. — The following table 
shows the values of the deflections under an arbitrary 
load P, or W, (within elastic limit), and of the safe load; 





Cantilevers. 


Beams with two end supports. 




With one end 
load P 
Fig. 241 


With unif . load 
W=wl 
Fig. 242 


Load Pin 
middle 
Fig. 234 


Unif. loaa 
W=wl 
Fig. 235 


Deflection 


' EI 


EI 


1 PI 3 
48 m EI 


5 TT/3 
384* EI 


| Safe load (from ?11 


Jl'I 


R'l 


i*l 


8*'I 


1 = M m ) 


le 


le 


le 


le 


Relative strength 


1 


2 


4 


8 

128 
5 


j Relative stiffness 
j under same load 


1 


7 3 


16 


( Relative stiffness 
"( under safe load 


1 


Va 


4 


16 
5 


j Max. shear = Jm,(and 
( location, 


P, (at wall) 


IF, (at wall) 


MP, (at supp). 


% W, (at supp\ 



also the relative strength, the relative stiffness (under the « 
same load), and the relative stiffness under the safe load, 
for the same beam. 

The max. shear will be used to determine the proper 
web-thickness for /-beams and " built-girders." The stu- 
dent should carefully study the foregoing table, noting 
especially the relative strength, stiffness, and stiffness 
under safe load, of the same beam. 

Thus, a beam with two end supports will bear a double 



278 MECHANICS OF ENGINEERING. 

load, if uniformly distributed instead of concentrated in 
the middle, but will deflect j£ more ; whereas with a given 
load uniformly distributed the deflection would be only 
5/% of that caused by the same load in the middle, provided 
the elastic limit is not surpassed in either case. 

251. R', etc. For Various Materials. — The formula^ = 3I m , 

e 

irom which in any given case of flexure we can compute 
the value of p m , the greatest normal stress in any outer 
element, provided all the other quantities are known, 
holds good theoretically within the elastic limit only. 
Still, some experimenters have used this formula for the 
rupture of beams by flexure, calling the value of p m thus 
obtained the Modulus of Rzipture, R. R may be found to 
differ considerably from both the T or C of § 203 with 
some materials and forms, being frequently much larger. 
This might be expected, since even supposing the relative 
extension or compression (i.e., strain) of the fibres to be 
proportional to their distances from the neutral axis as 
the load increases toward rupture, the corresponding 
stresses, not being proportional to these strains beyond the 
elastic limit, no longer vary directly as the distances from the 
neutral axis ; and the neutral axis does not pass through the 
centre of gravity of the section, necessarily. 

The following table gives average values for R, R', R", 

and E for the ordinary materials of construction. E, the 

modulus of elasticity for use in the f ormulse for deflection, 

' is given as computed from experiments in flexure, and is 

nearly the same as E t and E c . 

In any example involving R' } e is usually written equal 
to the distance of the outer fibre from the neutral axis, 
whether that fibre is to be in tension or compression ; 
since in most materials not only is the tensile equal to the 
compressive stress for a given strain (relative extension 
or contraction) but the elastic limit is reached at about 
the same strain both in tension and compression. 



FLEXUKE. SAFE LOADS. 279 

Table foe Use in Examples in Flexure. 



Timber. 



Cast Iron. 



Wro't Iron. 



Steel. 



Max. safe stress in outer fi 

bre =i?'(lbs. per sq. inch) 
Stress in outer fibre at Elas. j 

limit =7?''(lbs. per sq. in.) j 
" Modul. of Rupture " ) 

=i?=lbs. per sq. inch. ) 
^=Mod. of Elasticity, 

=lbs. per sq. inch. 



4,000 

to 

20,000 

1-000,000 
to 

,000 



) 1,000,C 
V to 

) 3,000,C 



6,000 in tens. 
12,000 in comp. 



40,000 



ir,oco ; coo 



12,000 



17.000 



50,000 



15,000 

to 
40,000 



ro,ooo 



25,000,030 



120,000 
Hard Steel. 
30,000,000 



In the case of cast iron, however, (see § 203) the elastic 
limit is reached in tension with a stress =9,000 lbs. per 
sq. inch and a relative extension of JjfL of one per cent., 
while in compression the stress must be about double to 
reach the elastic limit, the relative change of form (strain) 
being also double. Hence with cast iron beams, once 
largely used but now almost entirely displaced by rolled 
wrought iron beams, an economy of material was effected 
by making the outer fibre on the compressed side twice 
as far from the neutral axis as that on the stretched side. 
Thus, Fig. 243, cross-sections with unequal flanges were 
used, so proportioned that the centre of 
gravity was twice as near to the outer 
fibre in tension as to that in compression, 
i.e., e 2 =2e l ; in other words more material 
J is placed in tension than in compression. 
The fibre A being in tension (within elas- 
tic limit), that at B, since it is twice as far from the neu- 
tral axis and on the other side, is contracted twice as much 
as A is extended ; i.e., is under a compressive strain 
double the tensile strain at A, but in accordance with the 
above figures its state of stress is proportionally as much 
within the elastic limit as that of A. 

Steel beams are gradually coming into use, and may ul- 
timately replace those of wrought iron. 



L-, r-U* 






e 2 


N 




i 






r 






H 



Fig. ?43. 



280 MECHANICS OF ENGINEERING. 

The great range of values of R for timber is due nor 
only to the fact that the various kinds of wood differ 
widely in strength, while the behavior of specimens of 
any one kind depends somewhat on age, seasoning, etc., 
but also to the circumstance that the size of the beam un- 
der experiment has much to do with the result. The ex- 
periments of Prof. Lanza at the Mass. Institute of Tech- 
nology in 1881 were made on full size lumber (spruce), of 
dimensions such as are- usually taken for floor beams in 
buildings, and gave much smaller values of R (from 3,200 
to 8,700 lbs. per sq. inch) than had previously been ob- 
tained. The loading employed was in most cases a con- 
centrated load midway between the two supports. 

These low values are probably due to the fact that in 
large specimens of ordinary lumber the continuity of its 
substance is more or less broken by cracks, knots, etc., 
the higher values of most other experimenters having 
been obtained with small, straight-grained, selected pieces, 
from one foot to six feet in length. 

The value R' =12,000 lbs. per sq. inch is employed by 
the N. J. Iron and Steel Co. in computing the safe loads 
for their rolled wrought iron beams, with the stipulation 
that the beams (which are high and of narrow width) must 
be secure against yielding sideways. If such is not the 
case the ratio of the actual safe load to that computed with 
R'= 12,000 is taken less and less as the span increases. 
The lateral security referred to may be furnished by the 
brick arch-filling of a fire-proof floor, or by light lateral 
bracing with the other beams. 

252. Numerical Examples. — Example 1. — A square bar of 
wrought iron, 1*4 in. in thickness is bent into a circular 
arc whose radius is 200 ft., the plane of bending being par- 
allel to the side of the square. Eequired the greatest nor- 
mal stress p m in any outer fibre. 

Solution. From §§ 230 and 231 we may write 

— ==/_ . .-. p=eE^rp, i.e., is constant. 
p e 



FLEXURE. SAFE LOADS. 281 

For the units inch and pound (viz. those of the table in § 
251) we have e=% in., ^=2,400 in., and #=25,000,000 lbs. 
per sq. inch, and .*. 

p=p m =}£x 25,000,000-$- 2,400 = 7,812 lbs. per sq. in., 

which is quite safe. At a distance of }4 inch from the 
neutral axis, the normal stress is =[}4s-%]Pm = %Pm = 
5,208 lbs. per sq. in. (If the force-plane (i.e., plane of 
bending) were parallel to the diagonal of the square, e 
would =}4x 1-5V2 inches, giving J? m = [7,812x^/2 ] lbs. 
per sq. in.) § 238 shows an instance where a portion, OG, 
Fig. 231, is bent in a circular arc. 

Example 2. — A hollow cylindrical cast-iron pipe of radii 
3 ^and 4 inches is supported at its ends and loaded in 
middle (see Fig. 234). Kequired the safe load, neglecting 
the weight of the pipe. From the table in § 250 we have 
for safety 

P=4^ 
le 

From § 251 we put R' = 6,000 lbs. per sq. in.; and from § 
247 I— ^(r x 4 — r 2 4 ); and with these values, r 2 being =1, r l = 
4, e=r x =4, ~=-f- and Z=144 inches (the inch must be the 
unit of length since E'— 6,000 lbs. per sq. inch) we have 

P=4x6,000x^. -f- (256-150)-$- [144x4] ,\ P= 3,470 lbs. 

The weight of the beam itself is G= Vr, (§ 7), i.e., 

£ =;r ( ri 2_r 2 2 )Z r = 1.(16-12^)144x^=443 lbs. 

(Notice that y, here, must be lbs., per cubic inch). This 
weight being a uniformly distributed load is equivalent to 
half as much, 221 lbs., applied in the middle, as far as the 
strength of the beam is concerned (see § 250), .\ P must be 
taken =3,249 lbs. when the weight of the beam is consid- 
ered. 



282 MECHANICS OF ENGINEERING. 

Example 3. — A wrought-iron rolled I-beam supported 

at the ends is to be loaded uniformly Fig. 235, the span 

being equal to 20 feet. Its cross-section, Fig. 244, has a 

r--5^-{ depth parallel to the web of 15 

^f^ inches, a flange width of 5 inches. 

/ ] In the pocket book of the Trenton 

i '~\P Co. it is called a 15-inch light I- 

beam, weighing 150 lbs. per yard, 

fig. 244. with a moment of inertia =523. bi- 



15 



LA 



quad, inches about a gravity axis perpendicular to the 
web (i.e., when the web is vertical, the strongest position) 
and = 15 biq. in. about a gravity axis parallel to the web 
(i.e., when the web is placed horizontally). 
First placing the web vertically, we have from § 250, 

W Y = Safe load, distributed, =8^5. With #=12,000, 

7!=523, Z=24G inches, e x =7^ inches, this gives 
W x = [8 x 12,000 x 523] ~ [240 X-f] =27,893 lbs. 

But this includes the weight of the beam, £=20 ft. X^-lbs. 
= 1,000 lbs.; hence a distributed load of 26,902 lbs., or 13.45 
tons may be placed on the beam (secured against lateral 
yielding). (The pocket-book referred to gives 13.27 tons 
as the safe load.) 

Secondly, placing the web horizontal, 

^ 2 =8^- = [8xl2,000xl5]4-[240x 5 / 2 ]=^ of W 1 

or only about 1 / 12 of W x . 

Example 4. — Ee quired the deflection in the first case of 
Ex. 3. From § 250 the deflection at middle is 



d v 



WJ? _ 5 8BI, I 3 _ 5 E P 



384 EI, 384 le L EI X 48 E ^ 



PLEXUKE. SAFE LOADS. 283 

.-. ^=0.384 in. 
Example 5. — A rectangular beam of yellow pine, of width 
Z>=4 inches, is 20 ft. long, rests on two end supports, and is 
to carry a load of 1,200 lbs. at the middle ; required the 
proper depth h. From § 250 

le Z 12 * \h 

,\ h 2 =6Pl- : r4:B r b. For variety, use the inch and ton. For 
this system of units P=0.60 tons, i2'=0.50 tons per sq. in., 
7=240 inches and o— 4 inches. 

... 7i 2 =(6x0.6x240)-^(4x0.5x4)=108sq. in. \\ ^=10.4 in. 

Example 6. — Suppose the depth in Ex. 5 to be deter- 
mined by the condition that the deflection shall be = Y^ 
of the span or length. We should then have from § 250 

d= 1 l= l PP 
500 48 EI 

Using the inch and ton, with .Z7= 1,200,000 lbs. per sq. in., 
which = 600 tons per sq. inch, and 7=y i2 6A 3 , we have 

w= 500x0.60x240x240x12 . k in _ 

48x600x4 

As this is > 10.4 the load would be safe, as well. 

Example 7. — Eequired the length of a wro't iron pipe 

supported at its extremities, its internal radius being 2^ 

in., the external 2.50 in., that the deflection under its own 

weight may equal 1 / m of the length. 579.6 in. Ans. 

Example 8.— Fig. 245. The wall is 6 feet high and one 

foot thick, of common brick work 

(see § 7) and is to be borne by an 

/-beam in whose outer fibres no 



zzze 



T~T 



w 



— p greater normal stress than 8,000 

"E lbs. per sq. inch is allowable. If 

fig. 245. a number of I-beams is available, 



284 MECHANICS OF ENGINEERING. 

ranging in height from 6 in. to 15 in. (by whole inches), 
which one shall be chosen in the present instance, if their 
cross-seetions are Similar Figures, the moment of inertia of 
the 15-inch beam being 800 biquad. inches ? 

The 12-inch beam. Ans. 



SHEARING STRESSES IN FLEXURE. 

253. Shearing Stresses in Surfaces Parallel to the Neutral 
Surface. — If a pile of boards (see Fig. 246) is used to sup- 
port a load, the boards being free to slip on each other, it 
is noticeable that the ends overlap, although the boards- 




Fig. 24G. Fig. 347. 

are of equal length (now see Fig. 247) ; i.e., slipping has 
occurred along the surfaces of contact, the combina- 
tion being no stronger than the same boards side by 
side. If, however, they are glued together, piled as in the 
former figure, the slipping is prevented and the deflection 
is much less under the same load P. That is, the com- 
pound beam is both stronger and stiffer than the pile of 
loose boards, but the tendency to slip still exists and is 
known as the "shearing stress in surfaces parallel to the 
neutral surface." Its intensity per unit of area will now 
be determined by the usual " free-body " method. In Fig. 
248 let AN' be a portion, considered free, on the left of any 




N N 



£ 



Fio. 248. 



SHEAR IX FLEXURE. 



285 



section N', of a prismatic beam slightly bent under forces 
in one plane and perpendicular to the beam. The moment 
equation, about the neutral axis at N' t gives 



^— =M' ; whence p' ■ 
e 



M'e 
I 



(1) 



Similarly, with AN a,s a free body, NN* being —dx, 

±-—=31; whence p=—~ . 
e I 



(2) 



p and p' are the respective normal stresses in the outer 
fibre in the transverse sections N and N' respectively. 

Now separate the block NN\ lying between these two 
consecutive sections, as a free body (in Fig. 249). And 

^ PART OF J ' 





BART OF J 



Fig. 250. 



furthermore remove a portion of the top of the latter block, 
the portion lying above a plane passed parallel to the neu- 
tral surface and at any distance z" from that surface. This 
latter free body is shown in Fig. 250, with the system of 
forces representing the actions upon it of the portions taken 
away. The under surface, just laid bare, is a portion of a sur- 
face (parallel to the neutral surface) in which the above men- 
tioned slipping, or shearing, tendency exists. The lower por- 
tion (of the block NN') which is now removed exerted this 



286 MECHANICS OF ENGINEERING. 

rubbing, or sliding, force on the remainder along the under 
surface of the latter. Let the unknown intensity of this 
shearing force be X(per unit of area) ; then the shearing 
force on this under surface is =Xy"dx, (y",= oa in figure, 
being the horizontal width of the beam at this distance z" 
from the neutral axis of N') and takes its place with the 
other forces of the system, which are the normal stresses 

between , and portions of J and J', the respective 

\_z= z" < 

total vertical shears. (The manner of distribution of J 
over the vertical section is as yet unknown ; see next arti- 
cle.) 

Putting I (horiz. compons.) = in Fig. 250, we have 



/ e ^p'dF— f e Z -pdF—Xy"dx=0 

J z" e J z" e 

.:Xy"dx= P'—P f zdF 
e z n 

But from eqs. (1) and (2), p'— p = (}F—M)^=t- dM, 
while from § 240 dM = Jdx ; 

.:Xy"dx=^ fzdF-.X =* fldF .... (3) 

/ J z ii ly" e/,// 

as the required intensity per unit of area of the shearing 
force in a surface parallel to the neutral surface and at a 
distance z" from it. It is seen to depend on the " shear " J 
and the moment of inertia I of the whole vertical section; 
upon the horizontal thickness y" of the beam at the sur- 
face in question ; and upon the integral f zdF, 

z" 
which (from § 23) is the product of the area of that part of 
the vertical section extending from the surface in question to 
the outer fibre, by the distance of the centre of gravity of that 
part from the neutral surface. 



SHEAR IN FLEXURE. 



287 



It now follows, from § 209, that the intensity (per unit 
area) of the shear on an elementary area of the vertical 
cross section of a bent beam, and this intensity we may call 
Z, is equal to that X, just found, in the horizontal section 
which is at the same distance (z") from the neutral axis. 

254. Mode of Distribution of J, the Total Shear, over the Verti- 
cal Cross Section. — The intensity of this shear, ^(lbs. per 
sq. inch, for instance) has just been proved to be 



Z=X=J Tl CzdF 

hi t/ ..a 



(4) 



AS. 



iy 

To illustrate -this, required the 
value of Z two inches above the neu- 
tral axis, in a cross section close to 
the abutment, in Ex. 5, § 252. Fig. 
251 shows this section. From it we 
have for the shaded portion, lying 
above the locality in question, y" = 

4 inches, and C e ~ ' zdF = (area 
J z"= 2 

of shaded portion) X (distance of 
its centre of gravity from NA) = 
(12.8 sq. in.) x (3.6 in.) = 46.08 cubic inches. 

The total shear J = the abutment reaction = 600 lbs., 
while I = -L b¥ = £. x 4 x (10.4) 3 = 375 biquad. inches. 
Both J and /refer to the whole section. 




.Z= 



600x46.08 



18.42 lbs. per sq. in., 



375x4 

qui*e insignificant. In the neighborhood of the neutral 
axis, where z" — 0, we have y" = 4 and 

p e n zdF= r e zdF = 20.8 x 2.6=54.8, 
J z"=0 J 

wh_.e J and I of course are the same as before. Hence 



for 



," = 







288 



MECHANICS OF ENGINEEKING. 



Z^Z^= 21.62 lbs. per sq. in. 




Fig. 252. 



At the outer fibre sine* J* zdF=0, z" being = e, Z is = 



for a beam of any shape. 

For a solid rectangular section like the 
above, Z and z" bear the same relation to 
each other as the co-ordinates of the para- 
bola in Fig. 252 (axis horizontal). 

Since in equation (4) the horizontal 
thickness, y" 9 from side to side ef the sec- 
tion of the locality where Z is desired, 

occurs in the denominator, and since / e zdF 



increases as z' grows numerically smaller, the following 
may be stated, as to the distribution of J 3 the shear, in 
any vertical section, viz.: 

The intensity (lbs. per sq. in.) of the shear is zero at 
the outer elements of the section, and for beams of ordi- 
nary shapes is greatest where the section crosses the neu- 
tral surface. For forms of cross section having thin webs 
its value may be so great as to require special investiga- 
tion for safe design. 

Denoting by Z the value of Z&t the neutral axis, (which 
=X in the neutral surface where it crosses the vertica 
section in question) and putting the thickness of the sub- 
stance of the beam = b at the neutral axis, we have, 



j [ area above 

Z =X =-yt X< neutral axis 

-"to ( (or below) 



x 



the dist. of its cent, 
grav.from that axis 



(5) 



255. Values of Zo for Special Forms of Cross Section. — From 
the last equation it is plain that for a prismatic beam the 
value of Z Q is proportional to J, the total shear, and hence 
to the ordinate of the shear diagram for any particular 
case of loading. The utility of such a diagram, as obtain- 



SHEAR IN FLEXURE. 



289 



edin Figs. 234-237 inclusive, is therefore evident, for "by 
locating the greatest shearing stress in the beam it 
enables us to provide proper relations between the load- 
ing and the form and material of the beam to secure safety 
against rupture by shearing. 

The table in § 210 gives safe values which the j*-3— 
maximum Z in any case should not exceed. It is 
only in the case of beams with thin webs (see Figs. 
238 and 240) however, that Z is likely to need at- 
tention. 

For a Rectangle we have, Fig. 253, (see eq. 5, § 






Fig. 253. 



254) b =b, I= l / l2 bh\ and fzdF=y 2 b'h . j 



(Jo 



:.Z q =Xq=^ i i.e., =|. (total shear)-4-(whole area) 



Hence the greatest intensity of shear in the cross-section 
is ~ as great per unit of area as if the total shear were 
uniformly distributed over the section. 




Fig. 254. 



Fig. 235. 



e 



•-. 



■fj 



-Hbt-\ 



Fig. 256. 



Fig. 257. 



11 



For a Solid Circular section Fig. 254 



Ib eJo 



nr 2 4:r 4 J 



%7;r*.2r 2 3>r 3 tit 2 



[See § 26 Prob. 3]. 

For a Hollow Circular section (concentric circles) Fig. 
255, we have similarly, 



290 



MECHANICS OF ENGINEERING. 



z - J Y 'r* 4r 2 ~r 2 2 4r 2 l 

° ^(r/-r/)2(r-r 2 )L 2 ' 3- 2 * 3tt J 

_4 J(r^—ri) 

~~3 " 7r(r 1 4 — rf)^— r 2 ) 

Applying this formula to Example 2 § 252, we first have- 
as the max. shear J m = y£P =1,735 lbs., this being the abut- 
ment reaction, and hence (putting tt = (22 -4- 7)) 

v ™o^ 4x7xl735[64-42.8] OQ , „ 

A maX * = 3x22 [ 256-i50](4-5.3) - 294 ** p6r B * ^ 

which cast iron is abundantly able to withstand in shear- 
ing. 

For a Hollow Rectangular Beam, symmetrical about its 
neutral surface, Fig. 256 (box girder) 

7== J#(Mi 2 -iW ) = _3 _WjHV]_ 

The same equation holds good for Fig. 257 (I-beam with 
square corners) but then b 2 denotes the sum of the widths 
of the hollow spaces. 



256. 



Shearing Stress in the Web of an I-Beam. — It is usual to 
consider that, with I-beams (and box- 
beams) with the web vertical the shear J, 

| in any vertical section, is borne exclusively 
by the web and is uniformly distributed 

|~n — over its section. That this is nearly true 
may be proved as follows, the flange area 
being comparatively large. Fig. 258. Let 

~ F x be the area of one flange, and F that of 
the half web. Then since 



Fig. 258. 



/-4W+2**! (|)> 



SHEAR IN FLEXURE. 291 

(the last term approximate, y 2 \ being taken for the radi- 
us of gyration of Fi,) while 

C zdF=F 1 ^-{-F -j-, (the first term approx.) we have 
Jo 2 4 



J CzdF 

lb "Ift^Wo) ' wmc11 - b h 



if we write (2F 1 -{-F ) + (6^+2^)=^ • But b h is the. 
area of the whole web, .*. the shear per unit area at the 
neutral axis is nearly the same as if J were uniformly dis- 
tributed over the web. E. g., with F x = 2 sq. in., and F 
— 1 sq. in. we obtain Z = 1.07 (J-7-b h ). 

Similarly, the shearing stress per unit area at n, the 
upper edge of the web, is also nearly equal to J -±- b h (see 

eq., 4 (§254) for then f" f e ( zdF) 1 = F x .y 2 \ nearly, 

while I remains as before. 

The shear per unit area, then, in an ordinary I-beam is 
obtained by dividing the total shear Jhy the area of the 
web section. 

Example.— It is required to determine the proper 
thickness to be given to the web of the 15-inch wrought- 
iron rolled beam of Example 3 of §252, the height of web 
being 13 inches, with a safe shearing stress as low as 4000 
lbs. per sq. in. (the practice of the N. J. Steel and Iron 
Co., for webs), the web being vertical. 

The greatest total shear, J m , occurring at either support 
and being equal to half the load (see table §250) we have 
with b = width of web, 

Z max.= ^-; i.e. 4000 = J??^ ... b = 0.26 inches. 

Mo & xl3 



292 



MECHANICS OF EXGIXEEKIXGL 



(Units, inch and pound). The 15-inch light beam of the 
N. J. Co. has a web T / 2 inch thick, so as to provide for a 
shear double the value of that in the foregoing example. 
In the middle of the span Z d = 0, since J = 0. 

257. Designing of Riveting for Built Beams. — The latter are 
generally of the I-beam and box forms, made by riveting 
together a number of continuous shapes, most of the ma- 
terial being thrown into the flange members. E. g., in fig. 
259, an I-beam is formed by riveting together, in the 
manner shown in the figure, a "vertical stem plate " or 
web, four " angle-irons," and two " flange-plates," each of 





Fig. 259. 



Fig. 260. 



these seven pieces being continuous through the whole 
length of the beam. Fig 260 shows a box-girder. If the 
riveting is well done, the combination forms a single rigid 
beam whose safe load for a given span may be found by 
foregoing rules ; in computing the moment of inertia, how- 
ever, the portion of cross section cut out by the rivet 
holes must not be included. (This will be illustrated in 
a subsequent paragraph.) The safe load having been com- 
puted from a consideration of normal stresses only, and 
the web being made thick enough to take up the max. 
total shear, J m , with safety, it still remains to design the 
riveting, through whose agency the web and flanges are 
caused to act together as a single continuous rigid mass. 
It will be on the side of safety to consider that at a given 



SHEAR IN FLEXUBE. 293 

locality in the beam the shear carried by the rivets con- 
necting the angles and flanges, per unit of length of beam, 
is the same as that carried by those connecting the angles 
and the web ("vertical stem -plate"). The amount of this 
shear may be computed from the fact that it is equal to 
that occurring in the surface (parallel to the neutral sur- 
face) in which the web joins the flange, in case the web 
and flange were of continuous substance, as in a solid I- 
beam. But this shear must be of the same amount per 
horizontal unit of length as it is per vertical linear unit in 
the web itself, where it joins the flange ; (for from § 254 Z 
—X.) But the shear in the vertical section of the web, 
being uniformly- distributed, is the same per vertical linear 
unit at the junction with the flange as at any other part 
of the web section (§ 256,) and the whole shear on the ver- 
tical section of web = J, the " total shear " of that section 
of the beam. 

Hence we may state the following : 

The riveting connecting the angles with the flanges, (or 
the web with the angles) in any locality of a built beam, 
must safely sustain a shear equal to J on a horizontal length 
equal to the height of web. 

The strength of the riveting may be limited by the re- 
sistance of the rivet to being sheared (and this brings 
into account its cross section) or upon the crushing resist- 
ance of the side of the rivet hole in the plate (and this in- 
volves both the diameter of the rivet and the thickness of 
the metal in the web, flange, or angle.) In its practice the 
N. J. Steel and Iron Co. allows 7500 lbs. per sq. inch shear- 
ing stress in the rivet (wrought iron), and 12500 lbs. per 
sq. inch compressive resistance in the side of the rivet- 
hole, the axial plane section of the hole being the area of 
reference. 

In fig. 259 the rivets connecting the web with the angles 
are in double shear, which should be taken into account in 
considering their shearing strength, which is then double ;, 
those connecting the angles and the flange plates are in 



294 MECHANICS OF ENGINEERING. 



le shear. In fig. 260 (box-beam) where the beam is 
built of two webs, four angles, and two flange plates, all 
the rivets are in single shear. If the web plate is very 
high compared with its thickness, vertical stiffeners in the 
form of T irons may need to be riveted upon them lat- 
erally [see § 314]. 

Example. — A built I-beam of wrought iron (see fig. 259) 
is to support a uniformly distributed load of 40 tons, its 
extremities resting on supports 20 feet apart, and the 
height and thickness of web being 20 ins. and y 2 in. re- 
spectively. How shall the rivets, which are \ in. in di- 
ameter, be spaced, between the web and the angles which 
are also y 2 in. in thickness ? Keferring to fig. 235 we find 
that J = y^ W = 20 tons at each support and diminishes 
regularly to zero at the middle, where no riveting will there- 
fore be required. (Units inch and pound). Near a sup- 
port the riveting must sustain for each inch of length of 
l>eam a shearing force of (J -4- height of web) = 40000 -4- 
20 in. = 2000 lbs. Each rivet, having a sectional area of 
X n ( 7 A) 2 = 0.60 sq. inches, can bear a safe shear of 0.60 
X 7500 = 4500 lbs. in single shear, and .-. of 9000 lbs. in 
double shear, which is the present case. But the safe 
compressive resistance of the side of the rivet hole in 
either the web or the angle is only ^ in. x % in. X 12500 
= 5470 lbs., and thus determines the spacing of the rivets 
as follows : 

2000 lbs. -r- 5470 gives 0.36 as the number of rivets per 
inch of length of beam, i.e., they must be 1 -4- 0.36 = 2.7 
inches apart, centre to centre, near the supports; 5.4 inches 
apart at J^ the span from a support; none at all in the 
middle. 

However, " the rivets should not be spaced closer than 
2}£ times their diameter, nor farther apart than 16 times 
the thickness of the plate they connect," is the rule of the 
N. J. Co. 

As for the rivets connecting the angles and flange plates, 
being in two rows and opposite (ia pairs) the safe shear- 



FLEXURE. BUILT BEAM. 



295 



ing resistance of a pair (each in single shear) is 9,000 lbs., 
while the safe compressive resistance of the sides of the 
two rivet holes in the angle irons (the flange plate being 
much thicker) is = 10,940 lbs. Hence the former figure 
(9,000) divided into 2,000 lbs., gives 0.22 as the number of 
pairs of rivets per inch of length of the beam ; i.e., the 
rivets in one row should be spaced 4.5 inches apart, centre 
to centre, near a support ; the interval to be increased in 
inverse ratio to the distance from the middle of span, 
(bearing in mind the practical limitation just given). 

If the load is concentrated in the middle of the span, 
instead of uniformly distributed, J is constant along each 
half-span, (see fig. 234) and the rivet spacing must accord- 
ingly be made the same at all localities of the beam. 



SPECIAL PROBLEMS IN FLEXURE, 

258. Designing Cross Sections of Built Beams. — The last par- 
agraph dealt with the riveting of the various plates ; we 
now consider the design of the plates themselves. Take 
for instance a built I-beam, fig. 261 ; one vertical stem- 




Fig. 261. 



296 MECHANICS OF ENGINEERING, 

plate, four angle irons, (each of sectional area = A, re- 
maining after the holes are punched, with a gravity axis 
parallel to, and at a distance = a from its base), and two 
flange plates of width = h, and thickness = t Let the 
whole depth of girder = h, and the diameter of a rivet 
hole =f. To safely resist the tensile and compressive 
forces induced in this section by M m inch -lbs. {M m being 
the greatest moment in the beam which is prismatic) we 
have from § 239, 

M m = EI (1) 

e 

B! for wrought iron = 12,000 lbs. per sq. inch, e is = yi h 
while i, the moment of inertia of the compound section, 
is obtained as follows, taking into account the fact that 
the rivet holes cut out part of the material. In dealing 
with the sections of the angles and flanges, we consider 
them concentrated at their centres of gravity (an approx- 
imation, of course,) and treat their moments of inertia 

about N as single terms in the series CdF z 2 

(see § 85). The subtractive moments of inertia for the 
rivet holes in the web are similarly expressed ; let b = 
thickness of web. 

f 7 N for web = -^b (h—2tf—2b t' [|— t— a'] 2 
1 i" N for four angles = 4A fi — t — a] 2 
{ J N for two flanges = 2(b—2f) t (!^) 2 

the sum of which makes the i" N of the girder. Eq. (1) may 
now be written 

MJl =I* (2) 



2B' 

which is available for computing any one unknown quan- 
tity. The quantities concerned in 7 N are so numerous and 
they are combined in so complex a manner that in any 
numerical example it is best to adjust the dimensions , of 
the section to each other by successive assumptions and 



FLEXURE BUILT BEAM. 297 

trials. The size of rivets need not vary much in different 
cases, nor the thickness of the web-plate, which as used 
by the N. J. Co. is " rarely less than y or more than 5£ 
inch thick." The same Co. recommends the use of a 
single size of angle irons, viz., 3" X 3" X y 2 '\ for built 
girders of heights ranging from 12 to 36 inches, and also 
y in. rivets, and gives tables computed from eq. (2) for 
the proportionate strength of each portion of the com- 
pound section. 

Example. — (Units, inch and pound). A built I-beam 
with end supports, of span = 20 ft. = 240 inches, is to 
support a uniformly distributed load of 36 tons = 72,000 lbs. 
If ^ inch rivets are used, angle irons 3" X 3" X j£", ver- 
tical web y 2 " in thickness, and plates 1 inch thick for 
flanges, how wide (b = ?) must these flange-plates be ? 
taking h = 22 inches = total height of girder. 

Solution. — From the table in § 250 we find that the max. 
M for this case is y& Wl, where W — the total distributed 
load (including the weight of the girder) and I = span. 
Hence the left hand member of eq. (2) reduces to 

Wl h^ 72000 x 240 x 22 

16 " B> ~ 16 x 12000 '— 1980 

That is, the total moment of inertia of the section must 
be = 1,980 biquad. inches, of which the web and angles 
supply a known amount, since b Q = y, t = 1", t f = y, 
a' = 1^", A= 2.0 sq. in., a = 0.9 7 , and h = 22", are 
known, while the remainder must be furnished by the 
flanges, thus determining their width b, the unknown 
quantity. 

The effective area, A, of an angle iron is found thus : 
The full sectional area for the size given, = 3 X ^ + 
Z/4 X y = 2.75 sq. inches, from which deducting for two 
rivet holes we have 

^4=2.75— 2x^X^ = 2.0 sq. in. 

The value a = 0.90" is found by cutting out the shape 



298 MECHANICS OF ENGINEEKING. 



£3 



of two angles from sheet iron, thus : i i 

and balancing it on a knife edge. (The |" 

gaps left by the rivet holes may be ignored, 
without great error, in finding a). Hence, FlG . 261 a 
substituting we have 

I N for web =i, . y 2 x20 3 —2x}4 . ^ [8^] 2 =282.3 
I N for four angles = 4x2x[9.10] 2 =662.5 
I N for two flanges=2(5— |)xlx(10^) 2 =220.4(&— 1.5) 
.-. 1980= 282.3+662.5+(Z>— 1.5)220.4 
whence b = 4.6 + 1.5 = 6.1 inches 

the required total width of each of the 1 in. flange plates. 
This might be increased to 6.5 in. so as to equal the 
united width of the two angles and web. 

The rivet spacing can now be designed by § 257, and 
the assumed thickness of web, }4 in., tested for the max. 
total shear by § 256. The latter test results as follows : 
The max. shear J m occurs near either support and = 
y 2 JF= 36,000 lbs. .•., calling b' Q the least allowable thickness 
of web in order to keep the shearing stress as low as 4,000 
lbs. per sq. inch, 

b' x 20" x 4000 =36000 .-. 6' =0.45" 

showing that the assumed width of }4 in. is safe. 

This girder will need vertical stiffeners near the ends, 
as explained subsequently, and is understood to be sup- 
ported laterally. Built beams of double web, or box- 
form, (see Fig. 260) do not need this lateral support. 

259. Set of Moving Loads.— When a locomotive passes over 
a number of parallel prismatic girders, each one of which 
experiences certain detached pressures corresponding to 
the different wheels, by selecting any definite position of 
the wheels on the span, we may easily compute the reac- 
tions of the supports, then form the shear diagram, and 
finally as in § 243 obtain the max. moment, M my and the 



ELEXURE. MOVING LOADS. 



299 



max. shear J m , for this particular position of the wheels. 
But the values of M m and J m for some other position may 
be greater than those just found. We therefore inquire 
which will be the greatest moment among the infinite 
number of (M m )'s (one for each possible position of the 
wheels on the span). It is evident from Fig. 236 from the 
nature of the moment diagram, that when the pressures or 
loads are detached, the M m for any position of the loads, 
which of course are in this case at fixed distances apart, 
must occur under one of the loads (i.e. under a wheel). 
We begin .*. by asking : What is the position of the set of 
moving loads when the moment under a given wheel is 
greater than will occur under that wheel in any other po- 
sition? For example, in Fig. 262, in what position of the 



-/-(*- ah 



O^lOO" 




Fig 262. 

loads Pi, P 2 , etc. on the span will the moment M 2f i.e., 
under P 2 , be a maximum as compared with its value under 
P 2 in any other position on the span. Let B be the resultant 
of the loads which are now on the span, its variable distance 
from be = x, and its fixed distance from P 2 = a'\ while 
a, b, c, etc., are the fixed distances between the loads 
(wheels). For any values of x , as the loading moves 
through the range of motion within which no wheel of the 
set under consideration goes off the span, and no new 

wheel comes on it, we have Ri=---j- P, and the moment 

under P 2 

=M 2 =R,\l-(x~-a')-]— PJ)-Plb+c) 

M 2 =?-(l^^^a / x)—P,b~P i (b-{-c) (1) 



i.e. 



300 MECHANICS OF ENGINEERING. 

In (1) we have M 2 as a function of x, all the other quan- 
tities in the right hand member remaining constant as the 
loading moves; x may vary from x=a-\-a' to 
x=l— (c-\-b— a). For a max. M 2 , we put dM 2 -t-dx=0, i. e. 

~(l-2x~+a)=0 .'. x (for Max M 2 ) = y 2 l+y 2 a' 
(j 

(For this, or any other value of x, d 2 M 2 -7-dx 2 is negative, 
hence a maximum is indicated). For a max. 3I 2i then, R 
must be as as far {%a) on one side of the middle of the 
span as P 2 is on the other ; i.e., as the loading moves, the 
moment under a given wheel becomes a max. when that 
wheel and the centre of gravity of all the loads (then on 
the span) are equi -distant from the middle of the span. 

In this way in any particular case we may find the 
respective max. moments occurring under each of the 
wheels during the passage, and the greatest of these is the 
M m to be used in the equation M m —R f I—e for safe loading.* 

As to the shear J, for a given position of the wheels this 
will be the greatest at one or the other support, and 
equals the reaction at that support. When the load moves 
toward either support the shear at that end of the beam 
evidently increases so long as no wheel rolls completely 
over and beyond it. To find J" max., then, dealing with 
each support in turn, we compute the successive reactions 
at the support when the loading is successively so placed 
that consecutive wheels, in turn, are on the point of roll- 
ing off the girder at that end ; the greatest of these is the 
max. shear, J m . As the max. moment is apt to come under 
the heaviest load it may not be necessary to deal with 
more than one or two wheels in finding M m . 

Example. — Given the following wheel pressures, 

A< . . 8' . . >B< . . 5' . . >G< . . 4 . . <D 
4 tons. 6 tons. 6 tons. 5 tons, 

on one rail which is continuous over a girder of 20 ft. span, 
under a locomotive. 



* Since this may be regarded as a case of " sudden application" of a load, it is 
customary to make R' much smaller than for a dead load; from one-third to one-halt 

smaller. 



FLEXURE. MOVING LOADS. 



301 



1. Required the position of the resultant of A, B, and C » 

2. " " " " A, B, C, and I) ; 

3. " " " " B, C, and B. 

4. In what position of the wheels on the span will the 
moment under B be a max. ? Ditto for wheel C? Required 
the value of these moments and which is M m ? 

5. Required the value of J m , (max. shear), its location and 
the position of loads. 

Results.— (1.) 7.8' to right of A. (2.) 10' to right of A. 
(3.) 4.4' to right of B. (4.) Max. M B = 1,273,000 inch lbs. 
with all the wheels on ; Max. M G = 1,440,000 inch-lbs. with 
wheels B, C, and D on. (5.) J m = 13.6 tons at right sup- 
port with wheel B close to this support. 

260. Single Eccentric Load. — In the following special cases 
of prismatic beams, peculiar in the distribution of the 

loads, or mode of support, or both, 
the main objects sought are the 
^ values of the max. moment M mi for 
use in the equation 

i>i m =^ J (see§239); 
e 

and of the max. shear J m , from 

which to design the web riveting 

in the case of an I or box-girder, 

The modes of support will be such 

that the reactions are independent 

of the form and material of the 

fig. 263. beam (the weight of beam being 

neglected). As before, the flexure is to be slight, and the 

forces are all perpendicular to the beam. 

The present problem is that in fig. 263, the beam being 

prismatic, supported at the ends, with a single eccentric 

load, P. We shall first disregard the weight of the beam 

itself. Let the span =\-\-l 2 . First considering the whole 

beam free we have the reactions i? x = Pl 2 +• I and R 2 = 

Ph -4- I. 

Making a section at m and having Om free, x being < l 2 , 
2 (vert, compons.) = gives 




302 MECHANICS OF ENGINEERING. 

B 2 — e/=0, i.e., J=B 2 ; 
while from 2 (mom.) m =0 we have 

£ I -B,x= .\M = B 2 x=^x 

e I 

These values of J and M hold good between and (7, J 
being constant, while M is proportional to x. Hence for 
00 the shear diagram is a rectangle and the moment dia- 
gram a triangle. By inspection the greatest M for 00 is 
for x = l 2 , and = Bl^ 2 -f- I. This is the max. M for the 
beam, since between and B, M is proportional to the dis- 
tance of the section from B. 

, % j[f m =^ a nd^l i =4^ 2 . . . (1) 

lei 

is the equation for safe loading. 

J ' = R l in any section along CB, and is opposite in sign 
to what it is on 00; i.e., practically, if a dove-tail joint 
existed anywhere on 00 the portion of the beam on the 
right of such section would slide downward relatively to* 
the left hand portion ; but vice versa on OB. 

Evidently the max. shear J m = B x or B 2) as l 2 or \ is the 
greater segment. 

It is also evident that for a given span and given beam 
the safe load P', as computed from eq. (1) above, becomes 
very large as its point of application approaches a sup- 
port ; this would naturally be expected but not without 
limit, as the shear for sections between the load and the 
support is equal to the reaction at the near support and 
may thus soon reach a limiting value, when the safety of 
the web or the spacing of the rivets, if any, is considered. 

Secondly ', considering the weight of the beam, or any 
uniformly distributed loading, weighing w lbs. per unit of 
length of beam, in addition to P, Fig. 264, we have the 
reactions * 

*>= r M-- »< *•=?+? 

Let l 2 be >\\ then for a portion Om of length #<^ 
moments about m give 



FLEXURE. SPECIAL PROBLEMS. 



303 



X 



RoX -\-wx.~=0 



pi 

— ^ , _ a 

i.e., on OC, M=E 2 x — y 2 wx 2 

Evidently for x = (i.e. at 0) M = 0, while for a; 

at (7) we have, putting w = W -f- I 



Jf c =RA 



w?= 



PW 2 , Wk 



+- 



JPZ2 



(2) 
Z 2 (i.e. 

(3) 



« "2 '" I 
It remains to be seen whether a value of M may not exist 
in some section between and (7, (i.e., for a value of x 
<l 2 in eq. (2)), still greater than M c . Since (2) gives M as 
a continuous function of x between and C, we put 
dM-Sr dx = 0, an4 obtain, substituting the value of the con- 
stants B 2 and w, 

{ max. \ pti 



R 2 — wx=0 



W 



This must be for 
R 4 w=wj 

luiiiiii 



. x Q 1 for M or \ =~ l + % I 
( min. ) w 

M max., since d 2 M -*- dec 2 is negative 



Ri 




when this value of x is sub- 
stituted. If the particular 
value of x given by (4) is 
<l 2i the corresponding value 
of M (call it M n ) from eq. 
(2) will occur on OC and will 
be greater than M G (Dia- 
grams II. in fig. 264 show 



this 



case, 



but if x n is> l 2 



Fig. 264. 



we are not concerned with 
the corresponding value of 
M, and the greatest M on OC 
would be M c . 

For the short portion BC, 
which has moment and shear 
diagrams of its own not con- 
tinuous with those for OC, it 
may easily be shown that 
M c is the greatest moment of 
any section. Hence the M 



304 



MECHANICS OF ENGI^EEKING. 



max., or M m , of the whole beam is either M G or M n , 
according as x u is > or < l 2 . This latter criterion may be 
expressed thus, [with l 2 — }4 I denoted by Z 3 , the distance 
of P from the middle of the span] : 

and since from (4) and (2) 

PI, . . _i rA 
W 



From (eq. 4) 



IX. = 



Y+y 2 w^ 






The equation for safe loading is 



and 



nr /= **Hr ta> r; 



P'J 



P • 



= i!f n , when — - is < ~? 



JF 



?i 



. . . . (6) 

Seeeqs. (3) and (5) 
for M P and ilf n 



If either P, PT, l B , or Z x is the unknown quantity sought, the 
criterion of (6) cannot be applied, and we .*. use both equa- 
tions in (6) and then discriminate between the two results. 
The greatest shear is J m !Ss B l9 in Fig. 264, where l 2 is 

261. Two Equal Terminal Loads, Two Symmetrical Supports 
Tig. 265. [Same case as in Fig. 231, § 238]. Neglect 
weight of beam. The reaction at each support being =P, 
(from symmetry), we have for a free body Om with x < l x 



pi 



Px—^ = .-. M=Px 

e 



while where x>\ and < \-\-\ 

Px-p (x_^)_^ = o .-. M=P\ 



(i; 



(2) 



That is, see (1), ilff varies directly with x between and C, 
while between C and D it is constant. Hence for safe 
loading 

(3) 



— M 

e 



i.e., — ^Pli 



FLEXURE. SPECIAL PROBLEMS. 



305 



}P 






^ 



-I,— 



J-ZI 



-x n - 



jM 


Ijll^ 




MOMS. 






1IS11IHI1 


SHEARS 







Fig. 265 



The construction of the 

B moment diagram is evident 

p from equations (1) and (2). 

As for J> the shear, the 

same free bodies give, from 

2 (vert. forces)=0. 

On OG . J=P ... (4) 
On CD . J=P—P=zero\$) 



(4) and (5) might also be ob- 
tained from (1) and (2) by- 
writing J=d 31-7- dx, but the 
former method is to be preferred in most cases, since the 
latter requires M to be expressed as a function of x while 
the former is applicable for examining separate sections 
without making use of a variable. 

If the beam is an I-beam, the fact that e/is zero any- 
where on G D would indicate that we may dispense with 
a web along G D to unite the two flanges ; but the lower 
flange being in compression and forming a " long column " 
would tend to buckle out of a straight line if not stayed by 
a web connection with the other, or some equivalent brac- 
ing. 

262. Uniform Load over Part of the Span. Two End Supports. 
Pig. 266. Let the load= W, extending from one support 
over a portion = c, of the span, (on the left, say,) so that 
W= ivc, w being the load per unit of length. Neglect 
wei ght o f beam. For a free body Om of any length 
x < B (i.e. < c), I moms m =0 gives 



pi 



tax 



-BjX^O .\M=B 1 x- 



wx* 



which holds good for any section on B. As for sections 
on B C it is more simple to deal with the free body m'G, 
of length 



x 9 < C B from which we have M= R 2 x' 



(2) 



306 



MECHANICS OF ENGINEERING. 



13 




Fig. 266. 



which shows the moment 
curve for B G to be a straight 
line DC , tangent at D to the 
parabola 0' D representing 
eq. (1.) (If there were a con- 
centrated load at B, CD 
would meet the tangent at 
D at an angle instead of co- 
inciding with it ; let the stu- 
dent show why, from the 
shear diagram). 

The shear for any value of 
x on B is: 



On OB 
while on B C 



. J=B l —tvx . . . (3) 
. J= B 2 = constant . (4) 



The shear diagram is constructed accordingly. 

To find the position of the max. ordinate of the para- 
bola, (and this from previous statements concerning the 
tangent at the point D must occur on B, as will be seen 
and will .-. be the M m for the whole beam) we put J=0 in 
eq (3) whence 



x 



W IV I M 



(5) 



W 



and is less than c, as expected. [The value of B i== — Q — Q t 

=(^c +1) (I — £), (the whole beam free) has been substi- 
tuted]. This value of x substituted in eq. (1) gives 



HL 



a-y^r.*/. 



Wc 



B'l 



=m- 



4'^ywc... (6) 



is the equation for safe loading. 

The max. shear J m is found at and is = B» which is 
evidently >B 2 , at C. 



FLEXURE. SPECIAL PROBLEMS. 



307 



263. Uniform Load Ovex 4 Whole Length With Two Symmetrica, 
Supports. Fig. 267. — With the notation expressed in the fig- 
ure, the following results may be obtained, after having 
divided the length of the beam into three parts for sepa- 
rate treatment as necessitated by the external forces, which 
are the distributed load W, and 
and the two reactions, each = 
]/ 2 W. The moment curve is 
made up of parts of three dis- 
tinct parabolas, each with its 
axis vertical. The central par- 
abola may sink below the hori- 
zontal axis of 'reference if the 
supports are far enough apart, 
in which case (see Fig.) the elas- 
tic curve of the beam itself becomes concave upward be- 
tween the points E and F of " contrary flexure." At each 
of these points the moment must be zero, since the radius 
of curvature is oo and M = EI ■+ p (see § 231) at any sec- 
tion ; that is, at these points the moment curve crosses its 
horizontal axis. 

As to the location and amount of the max. moment M m9 . 
inspecting the diagram we see that it will be either at H, 
the middle, or at both of the supports B and G (which from 
symmetry have equal moments), i.e., 




Fig. 267. 



w. 



M m \ and.-. — \=\ * 

I -El? at B and C 



or 



according to which is the greater in any given case ; i.e. 
according as l 2 is > or < \ ^/§ t 

The shear close on the left of B = wl 1} while close to the 
right oi Bit= y 2 W - w\. (It will be noticed that in this, 
case since the beam overhangs, beyond the support, the: 
shear near the support is not equal to the reaction there, 
as it was in some preceding cases.) 



308 



MECHANICS OF ENGINEERING. 



Hence J„ 



{ y 2 w-wl, } accordin g as h ^A.l 



264, Hydrostatic Pressure Against a Vertical Plank. — From 
elementary hydrostatics we know that the pressure, per 
unit area, of quiescent water against the vertical side of a 
tank, varies directly with the depth, x, below the surface, 
and equals the weight of a prism of water whose altitude 
= x, and whose sectional area is unity. ^ee Fig. 268. 




Fig. 268. 

The plank is of rectangular cross section, its constant 
breadth, = b, being r~ to the paper, and receives no sup- 
port except at its two extremities, and B, being level 
with the water surface. The loading, or pressure, per unit 
of length of the beam, is here variable and, by above defini- 
nition, is = w = yxb, where r = weight of a cubic unit 
(i.e. the heaviness, see § 7) of water, and x = Om = depth 
of any section m below the surface. The hydrostatic pres- 
sure on dx = wdx. These pressures for equal dx's, vary 
as the ordinates of a triangle 0R x B. 

Consider Om free. Besides the elastic forces of the ex- 
posed section m, the forces acting are the reaction R , and 
the triangle of pressure OEm. The total of the latter is 



W li = f wdx —yb I xdx 






(i) 



and the sum of the moments of these pressures about m is 
equal to that of their resultant (= their sum, since they 

are parallel) about m, and .*. 



>*2 



FLEXURE. SPECIAL PROBLEMS. 30il 

[From (1) when x = Z, ve have for the total water pres- 

Z 2 
sure on the beam W l = yb -^ and since one-third of this 

will be borne at we have R Q =% ybl 2 .] 

Now putting I( moms, about the neutral axis of m)=0 r 
for Om free, we have 

B (fi c—JV x . ?— H=0 .-. 31= %yb (Vx—tf) '. (2) 

(which holds good from x = to x = I), From 1 (horiz. 
forces) = we have also the shear 

J=B U —W x =y 6 yb (Z 2 — 3^) .... (3) 

as might also have been obtained by differentiating (2), 
since J = dM -4- dx. By putting J = (§ 240, corollary) 
we have for a max. M, x = I -f- -v/3, which is less than I 
and hence is applicable to the problem. Substitute this 
in eq. 2, and reduce, and we have 

R'l _ . B'l 1 1 

— =Jl m ,i.e. _= fi .-^..,» .... (4) 

as the equation for safe loading. 

265. Example. — If the thickness of the plank is h, re- 
quired h = ?, if R' is taken = 1,000 lbs. per sq. in. for 
timber (§ 251), and I = 6 feet. For the inch-pound-second 
system of units, we must substitute R' = 1,000 ; I = 72 
inches ; y — 0.036 lbs. per cubic inch (heaviness of water 
in this system of units); while i" =bh? -f- 12, (§ 247), and e 
h. Hence from (4) we have 



= % 



1000 W 0.0366 x72 3 M ' , 07 . 

^ — -yr = n 7 - , .*. A 2 =5.16 .*. A = 2.27 in. 

It will be noticed that since x for M m = I -f- V 3, and not 
^3 Z, Jf m does not occur in the section opposite the resul- 
tant of the water pressure ; see Fig. 268. The shear curve 
is a parabola here ; eq. (3). 



310 



MECHANICS OF ENGINEERING. 



wdz '■ 



J.+ dJ 



ILL 



266. The Four x-Derivatives of the Ordinate of the Elastic Curve 
— If y = func. (x) is the equation of the elastic curve for 
any portion of a loaded beam, on which portion the load 
per unit of length of the beam is w = either zero, (Fig. 
234) or = constant, (Fig. 235), or = a continuous func. (x) 
(as in the last §), we may prove, as fol- 
lows, that w = the cc-derivative of the 
shear. Fig. 269. Let N and N' be two 
— * consecutive cross-sections of a loaded 

W"ik beam, and let the block between them, 

p- bearing its portion, wdx, of a distributed 
load, be considered free. The elastic 
forces consist of the two stress-couples 
(tensions and compressions) and the two 
shears, «7and J + dJ, dJbeing the shear-increment conse- 
quent upon x receiving its increment dx. By putting 
^(vert. components) = we have 



Fife 



Fig. 269. 



J-\-dJ- — ivdx — J=0 ,\ w= 



dJ 

dx 



Q. E. D. But J itself = dM ~ dx, (§ 240) and 

M = [d 2 y -T- dx 2 ] EI. By substitution, then, we have the 

following relations : 



y =func.(a?)= ordinate at any point of the elastic curve (1) 



■fy. 

dx 



a = slope at any point of the elastic curve 



(2) 



EI -r^~— M = ordinate (to scale) of the moment curve (3) 

dx % 

^-r d z v .i -L t \ the ordinate (to scale) ) /A s 

EI % = the ' i ° f the shear W«* I ■ • () 

( the load per unit of length \ 
EI ™M = w = \ of beam = ordinate (to scale) V . (5) 
dx* ( of a curve of loading. ) 

If, then, the equation of the elastic curve (the neutral line 
of the beam itself ; a reality, and not artificial like the 



FLEXURE. SPECIAL PROBLEMS. 311 

other curves spoken of) is given ; we may by successive 
differentiation, for a prismatic and homogeneous beam so 
that both E and / are constant, find the other four quan- 
tities mentioned. 

As to the converse process, (i.e. having given w as a 
function of x, to find expressions for J f M and y as func- 
tions of x) this is more difficult, since in taking the 
cc-anti-derivative, an unknown constant must be added and 
determined. The problem just treated in § 264, however, 
offers a very simple case since iv is the same function of 
x, along the whole beam, and there is therefore but one elas- 
tic curve to be determined. 

We .*. begin, numbering backward, with 

rrr d 4 V j, ( since w = rbx, see ) ,~ ^ 

M dtf = " T \ l^t § and Fig. 268 }. • • • ( 5a > 

[N. B. — This derivative (dJ-^-dx) is negative since dJ and 
dx have contrary signs.] 

,\ (shear=)E 7^/ _ r b ^+ Const 

dx 3 2 

But writing out this equation for x=0, i.e. for the point 
0, where the shear=i? , we have B = + Const. .*. Const. — 
R di and hence write 

EI^=— yb x l+R, . (Shear) . (4a) 
Again taking the cc-anti-derivative of both sides 

(Momenta) E I ^=— r b*L+R Q x+(Const.=()) . (3a) 
[At 0, x=0 also M, .-. Const. =0]. Again, 

At 0, where x=0 dy-±-dx=OQ = the unknown slope of the 
elastic line at 0, and hence C'=EI a^ 

■•■ EI %=-r h u+ R 4 +EIa « ■ • • (2a) 



S12 MECHANICS OF ENGINEERING. 

Passing now to y itself, and remembering that at 0, both 
y and x are zero, so that the constant, if added, would= 
zero, we obtain (inserting the value of i? from last §) 

EI y—r^+rbP^ + Eia oX . ( i«) 

the equation of the elastic curve. This, however, contains 
the unknown constant a =the slope at 0. To determine 
a Q write out eq. (la) for the point B, Fig. 268, where x is 
known to be equal to 7, and y to be = zero, solve for « , 
and insert its value both in (la) and (2a). To find the 
point of max. y (i.e., of greatest deflection) in the elastic 
curve, write the slope, i.e. dy -J- dx, = zero [see eq. 2a] and 
solve for x ; four values will be obtained, of which the one 
lying between and I is obviously the one to be taken. 
This value of x substituted in (la) will give the maximum 
deflection. The location of this maximum deflection is 



neither at the centre of action of the load 



(-SO 



nor at the section of max. moment (x =l-^\/ 3-) 

The qualities of the left hand members of equations (1) 
to (5) should be carefully noted. E. a., in the inch-pound- 
second system of units we should have : 

1. y (a linear quantity) = (so many) inches. 

2. dy-^-dx (an abstract number) == (so many) abstract 
units. 

3. M (a moment) = (so many) inch-pounds. 

4. «7(a shear, i.e., force) = (so many) pounds. 

5. w (force per linear unit) = (so many) pounds per run- 
ning inch of beam's length. 

As to the quantities E, and 7, individually, E is pounds 
per sq. in., and /has four linear dimensions, i.e. (so many) 
bi-quadratic inches. 



FLEXURE SPECIAL PROBLEMS. 313 

267. Resilience of Beam With End Supports.— Fig. 270. If a 
<?g mass whose weight is G (G large com- 

\ h , pared with that of beam) be allowed to 



r* 



fall freely through a height = h upon 

the centre of a beam supported at its 

frj *p*» extremities, the pressure P felt by the 

fig. 270. beam increases from zero at the first 

instant of contact up to a maximum P m , as already stated 

in §233a, in which the equation was derived, d m being 

small compared with h, 

1 P 2 7 3 

• °^k-¥i ■ ■ ■ w 

The elastic limit is supposed not passed. In order that 
the maximum normal stress in any outer fibre shall at most 
be=i2, a safe value, (see table §251) we must put 

= -j- [according to eq. (2) §241,] i.e. in equation (a) 

above, substitute P m = [4 B'l] -r-Ze, which gives 

73/277 r>/2 -u, j>n yi 

having put I=F¥ (k being the radius of gyration §85) 
and Fl= V the volume of the (prismatic) beam. From 
equation (b) we have the energy, Gh, (in ft. -lbs., or inch- 
lbs.) of the vertical blow at the middle which the beam of 
Fig. 270 will safely bear, and any one unknown quantity 
can be computed from it, (but the mass of G should not 
be small compared with that of the beam.) 

The energy of this safe impact, for two beams of the 
same material and similar cross-sections (similarly placed), 
is seen to be proportional to their volumes; while if further- 
more their cross-sections are the same and similarly 
placed, the safe Gh is proportional to their lengths. (These 
same relations hold good, approximately, beyond the elas- 
tic limit.) 

It will be noticed that the last statement is just the con* 



314 



MECHANICS OF ENGINEERING. 



Terse of what was found in §245 for static loads, (the 
pressure at the centre of the beam being then equal to 
the weight of the safe load) ; for there the longer the beam 
(and .*. the span) the less the safe load, in inverse ratio. 
As appropriate in this connection, a quotation will be 
given from p. 186 of " The Strength of Materials and 
Structures," by Sir John Anderson, London, 1884, viz.: 

" It appears from the published experiments and state- 
ments of the Bailway Commissioners, that a beam 12 feet 
long will only support ]/ 2 of the load that a beam 6 feet 
long of the same breadth and depth will support, but that 
it will bear double the weight suddenly applied, as in the 
case of a weight falling upon it," (from the same height, 
should be added) ; " or if the same weights are used, the 
longer beam will not break by the weight falling upon it 
unless it falls through twice the distance required to frac- 
ture the shorter beam." 

268. Combined Flexure and Torsion. Crank Shafts. Fig. 271. 
Let OiB be the crank, and NOi the portion projecting 

beyond the nearest bearing 
N. P is the pressure of the 
connecting-rod against the 
crank -pin at a definite in- 
stant, the rotary motion be- 
ing uniform. Let a= the 
perpendicular dropped from 
the axis 00 1 of the shaft 
upon P, and 1= the distance 
of P, along the axis Y from 
the cross-section NmN' of the 
Let NN be a diameter of this 
In considering the portion 
NO^B free, and thus exposing the circular section NmN', 
we may assume that the stresses to be put in on the ele- 
ments of this surface are the tensions (above NN') and 
the compressions (below NN') and shears ~| to NN', due 
to the bending action of P ; and the shearing stress tan- 




Fig. 5271. 



shaft, close to the bearing, 
section, and parallel to c 



FLEXURE. SPECIAL PROBLEMS. 



315 



gent to the circles which have as a common centre, and 
pass through the respective dF's or elementary areas, 
these latter stresses being due to the twisting action of P. 
In the former set of elastic forces let p = the tensile 
stress per unit of area in the small parallelopipedical ele- 
ment m of the helix which is furthest from NN' (the neu- 
tral axis) and 1= the moment of inertia of the circle about 
NN'\ then taking moments about NN' for the free body, 
(disregarding the motion) we have as in cases of flexure 
(see §239) 

pi «i.-. .. Plr 



PI ; i.e., p- 



(a) 



[None of the shears has a moment about M"'.] Next 
taking moments about 00 u (the flexure elastic forces, both 
normal and shearing, having no moments about 00,) we 
have as in torsion (§216) 



P*Iv -d • Par 

^—P=Pa; i.e., p = —^- 



(6) 



in which p s is the shearing stress per unit of area, in the 
torsional elastic forces, on any outermost dF, as at m ; 
and ip the polar moment of inertia of the circle about its 
centre 0. 

Next consider free,, in Fig. 272, a small parallelopiped 
taken from the helix at m (of Fig. 271.) The stresses [see 
§209] acting on the four faces |— to the paper in Fig. 272 
are there represented, the dimensions (infinitesimal) being 
n || to NJST, b || to 00,, and d 1 to the paper in Fig. 272. 



Pj>*~ 



n 



PM' 



pnd 



q#a 



Bi b 




qcd—f 



P s bdj 



pnd 



Fig. 272. 



Fiff. 273. 



316 MECHANICS OF ENGINEERING. 

By altering the" ratio of b to % we may make the angle $' 
what we please. It is now proposed to consider free the 
triangular prism, GST, to find the intensity of normal 
stress q, per unit of area, on the diagonal plane GH, (of 
length = c,) which is a bounding face of that triangular 
prism. See Fig. 273. By writing I (compons. in direc- 
tion of normal to GH)=-0, we shall have, transposing, 

qcd=pnd sin d+pjbd sin 0+p & nd cos 6 ; and solving for q 

71 [~b 71 

q=p~ sin d+p s -sin0+-, cos 6 J ; . (1) 

but n : c=sin 6 and b : c=cos d .'. 

q=p sin 2 0+p s 2 sin 6 cos . . (2) 

This may be written (see eqs. 63 and 60, O. W. J. Trigo- 
nometry) 

q=}4p(l-coB2d)+p a am2d . . (3) 

As the diagonal plane GH is taken in different positions 
(i.e., as 6 varies), this tensile stress q (lbs. per sq. in. for 
instance) also varies, being a function of 6, and its max. 
value may be >p. To find 6 for q max. we put 

g| , =p sin 2d+2p s cos 26, . . (4) 

2p 3 
= 0, and obtain : tan[2(# for q max)]= — — . . . (5) 

Call this value of 6, 0'. Since tan 26' is negative, 2d' lies 
either in the second or fourth quadrant, and hence 

*»*»-* v?tm andcos2 ^^fa (6) 

[See equations 28 and 29 Trigonometry, O. W. J.] The- 



FLEXUKE. CKANK SHAFT. 317 

upper signs refer to the second quadrant, the lower to the 
fourth. If we now differentiate (4), obtaining 

^=2j9cos20-4p 8 sin20 . . . (7) 

we note that if the sine and cosine of the [2$'] of the 2nd 
quadrant [upper signs in (6)] are substituted in (7) the re- 
sult is negative, indicating a maximum ; that is, q is a max- 
imum for 6= the d' of eq. (6) when the upper signs are taken 
(2nd quadrant). To find q max., then, put 6' for d in (3) 
substituting from (6) (upper signs). We thus find 

q max ==^[p+Vy+4p 3 2 .] . . (8) 

A similar process, taking components parallel to GH, 
Fig. 273, will yield q s max., i.e., the max. shear per unit of 
area, which for a given p and p s exists on the diagonal 
plane GH in any of its possible positions, as 6 varies. 
This max. shearing stress is 

? 8 max =^y5?+ip? . . (9) 

In the element diametrically opposite to m in Fig. 271, p 
is compression instead of tension ; q maximum will also 
be compression but is numerically the same as the q max. 
of eq. 8. 

269. Example.— -In Fig. 271 suppose P=2 tons = 4,000 
lbs., a=6 in., 1=5 in., and that the shaft is of wrought 
iron. Required its radius that the max. tension or com- 
pression may not exceed .£'=12,000 lbs. per sq. in.; nor the 
max. shear exceed >S" = 7,000 lbs. per sq. in. That is, we 
put ^=12,000 in eq. (8) and solve for r : also ^ s =7,000 in 
(9) and solve for r. The greater value of r should be 
taken. From equations (a) and (b) we have (see §§ 219 and 
247 for I p and 1) 



318 



MECHANICS OP ENGINEERING. 



m API , 2Pa 



which in (8) and (9) give 



max. q=y 2 ~ [4Z+V(4/)*+4(2a) 5 ] 



'tit* 



(8a) 



and 



max. g s =^^3A/(4Z) 2 +4(2a) 2 



(9a) 



With max. g= 12,000, and the values of P, a, and Z, already 
given, (units, inch and pound) we have from (8a), ^=2.72 
cubic inches .*. r=1.39 inches. 

Next, with max. q s = 7,000 ; P, a, and I as before ; from 
(9a), r 3 =2.84 cubic inches .*. r=1.41 inches. 

The latter value of r, 1.41 inches, should be adopted. It 
is here supposed that the crank-pin is in such a position 
(when P= 4,000 lbs., and a=6 in.) that q max. (and q 9 
max.) are greater than for any other position ; a number 
of trials may be necessary to decide this, since P and a are 
different with each new position of the connecting rod. If 
the shaft and its connections are exposed to shocks, R and 
>S" should be taken much smaller. 



270. Another Example of combined torsion and flexure is 
^ shown in Fig. 274. The 

\ p b work of the working force 

P 2 ( vertical cog-pressure) is 




Fig. 274. 



B expended in overcoming the 
resistance (another vertical 
cog-pressure) Q x . 

That is, the rigid body 
consisting of the two wheels and shaft is employed to 
transmit power, at a uniform angular velocity, and since 
it is symmetrical about its axis of rotation the iorces act- 
ing on it, considered free, form a balanced system. (See 
§ 114). Hence given P l and the various geometrical quan- 



FLEXURE. CRANK SHAFT. 319 

tities a lf b lf etc., we may obtain Q lf and the reactions P and 
P B , in terms of P x . The greatest moment of flexure in the 
shaft will be either P l lf at C ; or P B Z 3 , at JD. The portion 
CD is under torsion, of a moment of torsion = P 1 a 1 = Qfi^ 
Hence we proceed as in the example of § 269, simply put- 
ting iVi (or P^\i whichever is the greater) in place of PI, 
and P^ in place of Pa. We have here neglected the 
weight of the shaft and wheels. If Q { were an upward ver- 
tical force and hence on the same side of the shaft as P 1? 
the reactions P and P B would be less than before, and one 
or both of them might be reversed in direction. 



320 MECHANICS OF ENGINEERING. 



CHAPTER IV. 



FLEXURE, CONTINUED, 



CONTINUOUS GIRDERS, 



271. Definition. — A continuous, girder, for present pur- 
poses, may be defined to be a loaded straight beam sup- 
ported in more than two points, in which case we can no 
longer, as heretofore, determine the reactions at the sup- 
ports from simple Statics alone, but must have recourse 
to the equations of the several elastic curves formed by its 
neutral line, which equations involve directly or indirect- 
ly the reactions sought ; the latter may then be found as 
if they were constants of integration. Practically this 
amounts to saying that the reactions depend on the man- 
ner in which the beam bends ; whereas in previous cases, 
with only two supports, the reactions were independent of 
the forms of the elastic curves (the flexure being slight, 
however). 

As an Illustration, if the straight beam of Fig. 275 is placed 
on three supports 0, I>, and C, at the same level, the 
reactions of these supports seem at first sight indeterm- 
inate ; for on considering the p 1 h 

whole beam free, we have three j*— -~^ — 1 — "__^JP 

unknown quantities and only B /J-^ — p" ]Sp j-^t^ 

two equations, viz : S (vert. FlG . 275 

compons.) = and I (moms, about some point) = 0. If 
now be gradually lowered, it receives less and less pres- 



FLEXURE. CONTIGUOUS GIKDEKS. 321 

sure, until it finally reaches a position where the beam 
barely touches it ; and then O's reaction is zero, and B and 
C support the beam as if were not there. As to how 
low must sink to obtain this position, depends on the 
stiffness and load of the beam. Again, if be raised 
above the level of B and C it receives greater and greater 
pressure, until the beam fails to touch one of the other 
supports. Still another consideration is that if the beam 
were tapering in form, being stiffest at 0, and pointed at 
B and (7, the three reactions would be different from their 
values for a prismatic beam. It is therefore evident that 
for more than two supports the values of the reactions de- 
pend on the relative heights of the supports and upon the 
form and elasticity of the beam, as well as upon the load. 
The circumstance that the beam is made continuous over 
the support 0, instead of being cut apart at into two 
independent beams, each covering its own span and hav- 
ing its own two supports, shows the significance of the 
term " continuous girder." 

All the cases here considered will be comparatively 
simple, from the symmetry of their conditions. The 
beams will all be prismatic, and all external forces (i.e. 
loads and reactions) perpendicular to the beam and in the 
same plane. All supports at the same level, 

272. Two Equal Spans; Two Concentrated Loads, One in the Mid- 
dle of Each Span. Prismatic Beam. — Fig. 275. Let each half- 
span = y 2 l. Neglect the weight of the beam. Required 
the reactions of the three supports. Call them P B , P and 
P c . From symmetry P B = P c , and the tangent to the 
elastic curve at is horizontal ; and since the supports 
are on a level the deflection of C (and B) below O's tangent 
is zero. The separate elastic curves OB and DC have a 
common slope and a common ordinate at D. 

For the equation of OD, make a section n anywhere be- 
tween and D, considering nC a free body. Fig. 276 (a) 



322 



MECHANICS OF ENGINEERING. 




!T 



— H 






\- 



(fr 



Fig. 276. 



with origin and axis as there indicated. From I (moms 
about neutral axis of n) = we have (see § 231) 



EI % =nv*i-x)-?s-x) 



EI 



dx 



p(y 2 /x- 



X 2 ' 

*2 j 



-P c (lx— 



^)+(C=0) 



The constant = 0, for at both x, and dy ■*■ cfcc, = 0. 
Taking the x-anti-derivative of (2) we have 

'Ix 2 x\ -r.rlx 2 X s - 



^ =p (?-- 6 )- p {f-n 



(i 

(2) 
(3) 



Here again the constant is zero since at 0,x and y both =0. 
(3) is the equation of OD, and allows no value of x <0 
or>^. It contains the unknown force P c . 

For the equation of DC, let the variable section n be made 
anywhere between D and C, and we have (Fig. 276 (b) ; x 
may now range between y 2 l and T) 



dx 
dy 



-P c (lr-X) 



* 7 t=- p <HD +a ' 



(4) 



(5)' 



To determine (7', put x = y 2 l both in (5)' and (2), and 
equate the results (for the two curves have a common 
tangent line at D) whence C = }& PI 2 



El 



dy 
dx 



x 2 



y 8 pp-p c (ix-^ 



(5) 



FLBXUEB. CONTENTIOUS GIEDEES. 325 

Hence Ely = % PV&- P c Y&- °t] + C" . . (6)' 

At D the curves have the same y, hence put x = L in the 
right hand member both of (3) and (6)', equating results, 
and we derive C7"=— 1 PZ 3 

40 

EIy=^PVa>-P^_jt]^S.PW . . (6) 

which is the equation of DC, but contains the unknown 
reaction P c . To determine P c we employ the fact that O's 
tangent passes through C, (supports on same level) and 
hence when x = I in (6), y is known to be zero. Making 
these substitutions in (6) we have 

0=%pv-y iP ?-±pi? ... p c =±p 

From symmetry P B also = -^P, while P must = fP r 
since P B + P + P a = 2 P (whole beam free). [Note. — 
If the supports were not on a level, but if, (for instance) 
the middle support were a small distance = h^ below 
the level line joining the others, we should put x = I and 
y = — ho in eq. (6), and thus obtain P B = P c = ^ P + 

3EI~!, which depends on the material and form of the 
t 

prismatic beam and upon the length of one span, (whereas 
with supports aU on a level, P B = P c = A P is independent 
of the material and form of the beam so long as it is ho- 
mogeneous and prismatic.) If P , which would then = 
?! P — 6 EI (Jiq-^-W), is found to be negative, it shows that 
requires a support from above, instead of below, to 
cause it to occupy a position Jiq below the other supports, 
i.e. the beam must be " latched down " at 0.] 

The moment diagram of this case can now be easily con- 
structed ; Fig. 277. For any free body nG, n lying in DC r 
we have 



324 



MECHANICS OF ENGINEERING. 




Fig. 277. 

figure) and at 0, where x= I, becomes 

■■- M= 



i.e., varies directly as x, un- 
c til x passes D when, for any 
point on DO, 

M=%Px-P{x- 1 -) 

which is =0, (point of in- 
flection of elastic curve) 
for x= 8 / u I (note that x is 
measured from G in this 
PI 



|PZ; M G =0; M B = 3 \Pl; and Jf o «0 

Hence, since M max. =^Pl 3 the equation for safe loading 
is 

R' I 6 tvi /y\ 



AP, while on DO 



e 32^ ' 

The shear at and anywhere on CD= 
it =^P in the opposite direction 

.-.J m -5P . . • • (8) 

The moment and shear diagrams are easily constructed, 
as shown in Fig. 277, the former being svmmetrical about 
a vertical line through 0, the latter about the point 0" 
Both are bounded by right lines. 

273. Two Equal Spans. Uniformly Distributed Load Over 

Whole Length. Prismatic Beam. 
y ^ ^JOX* N c —Fig. 278. Supports B, 0, 

£ij I iJ J 1 1 U jljll G > on a level Total load 

f>^_T^- — oT~T?-- -""~t == ^ W= 2wl and may include 

|p ! w( > l : x > J that of the beam ; w is con- 
I I 1 1 1 I 1 ! stant. As before, from sym- 
metry P B =P cf the unknown 
reactions at the extremi- 
ties. 



Pb 



I 

Fig. 278. 




FLEXURE. CONTINUOUS GIRDERS. 325 

Let On=x ; then with :iG free, I 1 moms, about n=0 gives 



J-Xs 



IV 



EiyL=vi!-x)qfy-Pii-x)= £[P— »H-rf]--P,(H0 (i> 



— wyr-njy — , i 

. EI dy = w [Vx _ M + ^_ p ^ Jx _ ^ ]+[ Const=0] (2) 
ctx A o A 

[Const. =0 for at both dy-z-dx the slope, and x, are =0] 

... EIy= ^[jiW-^^+y u .^]-P o [jifa?-^]+(O=0) (3) 

A 

[Const. =0 for at both x and y are =0]. Equations (1), 
(2), and (3) admit of any value of x from to I, i.e., hold 
good for any point of the elastic curve 0(7, the loading on 
which follows a continuous law (viz. : w= constant). But 
when x—l, i.e., at (7, y is known to be equal to zero, since 
0, B and are on the axis of X, (tangent at 0). With 
these values of x and y in eq. (3) we have 

o= «£. . t-}iPj? .-. p c =3/ 8 id=y 8 w 



.-.p, 



^TFandP =2JF— 2P C - 



? JF 



The Moment and Shear Diagrams can now be formed since 

all the external forces are 
w known. In Fig. 279 meas- 
ure x from G. Then in any 
section n the moment of the 
"stress-couple " is 




M=y 8 Wx- 



wx 2 



2 



(1) 



which holds good for any 
value of x on CO, i.e., from 
x=0 up to x=l. By inspec- 
pig. 279. tion it is seen that for x=0, 

M=0 ; and also for x=ffl, M=0, at the inflection point G, 
beyond which, toward 0, the upper fibres are in tension 



326 MECHANICS OF ENGINEERING. 

the lower in compression, whereas between C and G they 
are vice versa. As to the greatest moment to be found on 
CG, put dM^-dx—0 and solve for x. This gives 

}i W—wx=0 .*. [x for M max.]=^ . (2) 

which in eq. (1) gives 

Jf H (atJViseefigure)=+£JPZ . . (2) 

But this is numerically less than M (= — 1 /& Wl) hence the 
stress in the outer fibre at being 

Po=}6- T » ... (3) 

the equation for safe loading is 

^L=y6Wl .... (4) 

the same as if the beam were cut through at 0, each half, 
of length I, retaining the same load as before [see § 242 eq. 
(2)]. Hence making the girder continuous over the mid- 
dle support does not make it any stronger under a uni- 
formly distributed load ; but it does make it considerably 
stiffer. 

As for the shear, J, we obtain it for any section by tak- 
ing the x-derivative of M in eq. (1), or by putting ^(ver- 
tical forces) =0 for the free body nC, and thus have for 
any section on GO 

J=y 8 W—wx ... (5) 

«7is zero for x = }£l (where M reaches its calculus maxi- 
mum M N ; see above) and for x=l it = — $i W which is nu 
merically greater than }i W, its value at 0. Hence 

J m = s/ 8 W .... (6) 



FLEXURE. CONTINUOUS GIRDERS. 327 

The moment curve is a parabola (a separate one for each 
span), the shear curve a straight line, inclined to the hor- 
izontal, for each span. 

Problem. — How would the reactions in Fig. 278 be 
changed if the support were lowered a (small) distance 
h Q below the level of the other two ? 

274. Prismatic Beam Fixed Horizontally at Both Ends (at 
Same Level). Single Load at Middle. — Fig. 280. [As usual 

j x~n p-j the beam is understood to 

f^ - — ^ \^ p J _ -— — ^~ _j be homogeneous so that E 

E »[ 1 — *** tf is the same at all sections]. 

t i The building in, or fixing, 

j j/j A of the two ends is supposed 

o Sj^Jl^ — — — y -— ^=^ — jc* to be of such a nature as to 

YJ /<>w ~ B r — --— ~^- z ~i cause no horizontal con- 
fig. 280. straint ; i.e., the beam does 

not act as a cord or chain, in any manner, and hence the 
sum of the horizontal components of the stresses in any 
section is zero, as in all preceding cases of flexure. In 
other words the neutral axis still contains the centre of 
gravity of the section and the tensions and compressions 
are equivalent to a couple (the stress-couple) whose mo- 
ment is the " moment of flexure." 

If the beam is conceived cut through close to both wall 
faces, and this portion of length=£, considered free, the 
forces holding it in equilibrium consist of the downward 
force P (the load) ; two upward shears J and J c (one at 
each section) ; and two " stress-couples " one in each sec- 
tion, whose moments are M andi!f c . From symmetry we 
know that J =J C , and that M -=M C . From 2 Y=0 for the 
free body just mentioned, (but not shown in the figure), 
and from symmetry, we have J = *4 P and J c = }4 P '■> but 
to determine M Q and M Q , the form of the elastic curves 
B and B G must be taken into account as follows : 
Equation of OB, Fig. 280. 2 [mom. about neutral axis 
of any section n on B] = (for the free body nC which 



328 MECHANICS OF ENGINEERING. 

has a section exposed at each end, n being the variable, 
section) will give 

EI %= P{1/2 i - x )+ M >-y> F «- x ) • • w 

[Note. In forming this moment equation, notice that 
M c is the sum of the moments of the tensions and com- 
pressions at about the neutral axis at n, just as much as 
about the neutral axis of 0; for those tensions and com- 
pressions are equivalent to a couple, and hence the sum of 
their moments is the same taken about any axis whatever 
"| to the plane of the couple (§32).] 

Taking the a>anti-derivative of each member of (1), 

Eip=P(y 2 ix-y 2 x>)+3fcX-y 2 p(ix-y 2 x>) . (2) 

a x 

(The constant is not expressed, as it is zero). Now from 
symmetry we know that the tangent-line to the curve B 
at B is horizontal, i.e., for x=yi, dy-7-dx=0, and these 
values in eq. (2) give us 

0=y PZ 2 + i/ 2 M c l—^Pl 2 ; whence 3I=M =y PI . (3) 

Safe Loading. Fig. 281. Having now all the forces which 
act as external forces in straining the beam 00, we are 
ready to draw the moment diagram and find M m . For con- 
venience measure x from (7. For the free body nO, we 
have [see eq. (3)] 

y 2 Px-m c +pJ=o .-. 3i=y pi-y 2 Px ... (4) 

1 ±p Eq. (4) holds good for any 
2 J section on OB. By put- 
»"* ting #=0 we have 3f=M c = 
PI; lay o$HO'=M c to 
scale (so many inch-pounds 
moment to the inch of pa- 
per). At B y for x=y I, 
M B = — y PI ; hence lay 
off B'D=y PI on the op- 
Fig. 281. posite side of the axis O'C 




FLEXURE. CONTINUOUS GIRDERS. 329 

from HC, and join DR. JDK, symmetrical with DH about 
B'B, completes the moment curves, viz.: two right lines. 
The max. M is evidently =y& PI and the equation of safe 
loading 

— =}iPl (5) 

Hence the beam is twice as strong as if simply supported 
at the ends, under this load ; it may also be proved to be 
four times as stiff. 

The points of inflection of the elastic curve are in the 
middles of the half-spans, while the max. shear is 

j m =y 2 p ....... (6) 

275. Prismatic Beam Fixed Horizontally at Both Ends [at Same 
Level], Uniformly Distributed Load Over the Whole Length. 
Fig. 282. As in the preceding problem, we know from 
symmetry that J =J C =}4 W— ]/ 2 wl, and that 3f =M c , and 
determine the latter quantities by the equation of the 
curve OG, there being but one curve in the present in- 
stance, instead of two, as there is no change in the law of 
loading between and O. With nO free, 2 (mom n )=0 
gives 

Ell^=-}4Wx + ]U +l%* . . . . (1) 

dx 2 2 v ' 

and.-. EI^=—y 2 W^ + M o x+^ + [C=0] . . (2) 



w=wl 




Fig. 282. 



330 



MECHANICS OF ENGINEERING. 



The tangent line at being horizontal we have for x=0,-j = 

0, .-. (7=0. But since the tangent line at C is also hori- 
zontal, we may for x—l put dy+dx=0, and obtain 



0=— %WP+m+j4wP; whence M =LJVl 



(3) 



as the moment of the stress-conple close to the wall at 
and at C. 

Hence, Fig. 283, the equation of the moment curve (a 
single continuous curve in this case) is found by putting 
I (mom n )=0 for the free body nO, of length x, thus 
obtaining 



W=wZ 



LUil I iLUil 







i.e. 



e 
M 



Fig. 283. 



Wx-31 



wx 2 



iwi+ 



ivxr 



Wx 



(4) 



an equation of the second degree, indicating a conic. At 0, 
M=M of course,= I- Wl ; at B by putting x=y 2 l in (4), we 
have M B =—i 4: Wl, which is less than M , although 3I B is the 
calculus max. (negative) for M, as may be shown by writ- 
ing the expression for the shear {J=*% W—wx) equal to 
zero, etc. 



FLEXUKE. CONTINUOUS GIRDERS. 331 

Hence M m =~Wl, and the equation for safe loading is 

^r=k wi (5) 

Since (with this form of loading) if the beam were n6t 
built in but simply rested on two end supports, the equa- 
tion for safe loading would be [B'I-*-e] = }£ Wl, (see §242), 
it is evident that with the present mode of support it is 50 
per cent, stronger as compared with the other ; i.e., as re- 
gards normal stresses in the outer elements. As regards 
shearing stresses in the web if it has one, it is no stronger* 
since J m =% Win both cases. 

As to stiffness under the uniform load, the max. deflec- 
tion in the present case may be shown to be only h of that 
in the case of the simple end supports. QUE 

It is noteworthy that the shear diagram in Fig. 283 is 
identical with that for simple end supports §242, under 
uniform load ; while the moment diagrams differ as fol- 
lows : The parabola KB' A, Fig. 283, is identical with that 
in Fig. 235, but the horizontal axis from which the ordi- 
nates of the former are measured, instead of joining the 
extremities of the curve, cuts it in such a way as to have 
equal areas between it and the curve, on opposite sides 

i.e., areas [KC'IP+A G'0']= area H'G'B' 

In other words, the effect of fixing the ends horizontally 
is to shift the moment parabola upward a distance = M c 
(to scale), = 1 Wl, with regard to the axis of reference, 
O'B', in Fig. 235. 

276. Remarks. — The foregoing very simple cases of con- 
tinuous girders illustrate the means employed for deter- 
mining the reactions of supports and eventually the max. 
moment and the equations for safe loading and for deflec- 
tions. When there are more than three supports, with 
spans of unequal length, and loading of any description, 
the analysis leading to the above results is much more 
complicated and tedious, but is considerably simplified 



332 MECHANICS OF ENGINEERING. 

and systematized by the use of the remarkable theorem of 
three moments, the discovery of Clapeyron, in 1857. By 
this theorem, given the spans, the loading, and the vertical 
heights of the supports, we are enabled to write out a rela- 
tion between the moments of each three consecutive sup- 
ports, and thus obtain a sufficient number of equations to 
determine the moments at all the supports [p. 641 Bankine's 
Applied Mechanics.] From these moments the shears 
close to each side of each support are found, then the 
reactions, and from these and the given loads the moment 
at any section can be determined ; and hence finally the 
max. moment M m9 and the max. shear J m . 

The treatment of the general case of continuous girders 
by graphic methods, however, is comparatively simple, 
and its presentation is therefore deferred, § 391. 



THE DANGEROUS SECTION OF NON-PRIS- 
MATIC BEAMS. 

277. Remarks. By " dangerous section " is meant that sec- 
tion (in a given beam under given loading with given mode 
of support) where p, the normal stress in the outer fibre, 
at distance e from its neutral axis, is greater than in the 
outer fibre of any other section. Hence the elasticity of 
the material will be first impaired k$ the outer fibre of 
this section, if the load is gradually increased in amount 
(but not altered in distribution). 

In all preceding problems, the beam being prismatic, I r 
the moment of inertia, and e were the same in all sections, 

hence when the equation £-=M [§239] was solved for p, 

e 

. . Me /1N 

giving P== ~T * * * ' ^ ' 

we found that p was a max., = p m , for that section whose 
M was a maximum, since p varied as M, for the moment 



FLEXURE NOX-PRISMATIC BEAMS. 333 

of the stress-couple, as successive sections along the beam 
were examined. 

But for a non-prismatic beam /and e change, from sec- 
tion to section, as well as M, and the ordinate of the 
moment diagram no longer shows the variation of p, nor 
is p a max. where M is a max. To find the dangerous 
section, then, for a non-prismatic beam, we express the M, 
the I, and the e of any section in terms of x, thus obtain- 
ing j9=func. (x), then writing dp-±-dx=0, and solving for x. 

278. Dangerous Section in a Double Truncated Wedge. Two 

End Supports. Single Load in Middle. — The form is shown in 

Fig. 284. Neglect weight of beam ; measure x from one sup- 

I ,.i port 0. The 

t\p jOX^^—--^ Y -e ~^ ! I reaction at 

Ix^^--^t2-~-4i1/ ^P^-^i ~*"|Fl23* eca k su PP or ^ 

A^^° \iT~t^^S^ | isJ^P. The 

. rn-^ ^^r^rr^?!!^ 11 " width of the 

Fig. 284. beam = & at 

all sections, while its height, v, varies, being = h at 0. 
To express thee = ]/ 2 v, and the I = 1 fo; 3 (§247) of any 
section on 0(7, in terms of x, conceive the sloping faces 
of the truncated wedge to be prolonged "to their intersec- 
tion A, at a known distance = c from tlieface at 0. We 
then have from similar triangles 

v : x + c : : h : c, .-. v = - (x + c) . . (1) 

c 

and .\ e = # * fo+c) and 7 = 1 & -[cc+c] 3 . (2) 
c 12 c s 

For the free body nO, Z (moms. n ) = gives 

y 2 Px-Pl=o,. P ~^. . . ..(3) 

[That is, the M = y 2 P x .] But from (2), (3) becomes 
By putting dp ■+ dx = we obtain both sc = — c, and 



334 



MECHANICS OF ENGINEERING. 



x = + c, of which the latter, x = + c, corresponds to a 
maximum for p (since it will be found to give a negative 
result on substitution in d 2 p -f- dx 2 ). 

Hence the dangerous section is as far from the support 
0, as the imaginary edge, A, of the completed wedge, but 
of course on the opposite side. This supposes that the 
half -span, *4l, is > c; if not, the dangerous section will 
be at the middle of the beam, as if the beam were 
prismatic. 

Hence, with [the equation for safej RW2 ' 



) at middle) 






(5). 



while with ) the equation for safe j m ^ y 

i/7 ^ „ /-loading is: (put x=c < i — -■ 

y* l>c [and ^=£'^[3]) ( 6 

(see §239.) 



Pc (6) 



279. Double Truncated Pyramid and Cone. Fig. 285. For 




Fig. 285. 



the truncated pyramid both width = u, and height = v, 
are variable, and if b and h are the dimensions at 0, and 
c = ~0A = distance from to the imaginary vertex A, we 

shall have from similar triangles u=^ (a?+c)and v— — (x+c). 

c c 

Hence, substituting e=j4v and 7=1 uv 3 , in the moment 
equation 



e 2 > bh 2 ( x +tf 



dx bh 



l_ ( x+cf—Sx (x+cf 
(x+cf 



(7) 

(8)- 



FLEXUKE. NON-PRISMATIC BEAMS. 335 

Putting this = 0, we have x = — c, x = — c, and a; = 
+ ]/ 2 c, hence the dangerous section is at a distance a? = j^ c 
from 0, and the equation for safe loading is 

either B ' b ' h ^^ PI . . . . if % Z is < % c .... (9) 
o 

(in which V and A' are the dimensions at mid-span) 

or B(*/ % b)C>/ 2 hY = y, pc if ^ ^ > ^ c ^ ^ ^ (1Q) 

For the truncated cone (see Fig. 285 also, on right) where 
e = the variable radius r, and / = j^ tz r 4 , we also have 

' /KConst.] .^ . . . . (11) 

and hence p is a max. for x = j£ c, and the equation for 
safe loading 

either 5£iZL = % Pl,ior %l < % c (12) 

(where r' = radius of mid-span section) ; 

or *fl' (l r o) 3 = ^ Pc, for iX Z> J 

4 
(where r = radius of extremity.) 



or **' (l r o) 3 = % Pc, for iX I > i/ c (13) 

4 



NON-PRISMATIC BEAMS OF "UNIFORM 
STRENGTH." 

280. Remarks. A beam is said to be of " uniform 
strength " when its form, its mode of support, and the dis- 
tribution of loading, are such that the normal stress p has 
the same value in all the outer fibres, and thus one ele- 
ment of economy is secured, viz. : that all the outer fibres 
may be made to do full duty, since under the safe loading, 
p will be = to R' in all of them. [Of course, in all cases 
of flexure, the elements between the neutral surface and 



336 



MECHANICS OF ENGINEERING. 



the outer fibres being under tensions and compressions 
less than B' per sq. inch, are not doing full duty, as 
regards economy of material, unless perhaps with respect 
to shearing stresses.] In Fig. 265, §261, we have already 
had an instance of a body of uniform strength in flexure, 
viz. : the middle segment, CD, of that figure ; for the 
moment is the same for all sections of CD [eq. (2) of that 
§], and hence the normal stress p in the outer fibres (the 
beam being prismatic in that instance). 

In the following problems the weight of the beam itself 
is neglected. The general method pursued will be to find 
an expression for the outer-fibre-stress p, at a definite sec- 
tion of the beam, where the dimensions of the section are 
known or assumed, then an expression for p in the varia- 
ble section, and equate the two. For clearness the figures 
are exaggerated, vertically. 

281. Parabolic Working Beam. Unsymmetrical. Fig. 286 

t; 




CBO is a working beam or lever, B being the fixed fulcrum 
or bearing. The force P being given we may compute P c 
from the mom. equation P l = P c l it while the fulcrum 
reaction is P B =P +P C . All the forces are ~| to the beam. 
The beam is to have the same width b at all points, and is 
to be rectangular in section. 

Kequired first, the proper height h^ at B, for safety. 



From the free body BO, of length 
= ; i.e., 



l oi we have 2 (moms B ) 



PbT_ 



PX 



ovp B 



6 P l 
bh 2 



(1) 



FLEXUKE. NON-PRISMATIC BEAMS. 337 

Hence, putting p B = R', \ becomes known from (1). 

Bequired, secondly, the relation between the variable 
height v (at any section n) and the distance x of n from 0. 
For the free body nO, we have (I moms n = 0) 

PJ= PqX . or Pn^W =PqX and ... p = ^ (2) 

e y 2V bv 2 

But for " uniform strength " p a must = p B ; hence 
equate their values from (1) and (2) and we have 

5 = A, which may be written (V 2 v) 2 = O^Z. x (3) 
v 2 h 2 l 

so as to make the relation between the abscissa x and the 
ordinate j4 v more marked ; it is the equation of a para- 
bola, whose vertex is at 0. 

The parabolic outline for the portion BC is found simi- 
larly. The local stresses at C, B, and must be proper- 
ly provided for by evident means. The shear J = P , at 
0, also requires special attention. 

This shape of beam is often adopted in practice for the 
working beams of engines, etc. 

The parabolic outlines just found may be replaced by 
trapezoidal forms, Fig. 287, without using much more ma- 
terial, and by making the slop- — ^^/^^^^-^ 
ing plane faces tangent to the ^^f^^f^^-^!^^ ^^] 

parabolic outline at points T 5^"~~~^ r ^^~~^~^T 

and T ly half -way between and ^ & 

B, and C and B, respectively. It fig. 287. 

can be proved that they contain minimum volumes, among 
all trapezoidal forms capable of circumscribing the given 
parabolic bodies. The dangerous sections of these trape- 
zoidal bodies are at the tangent points T and T x . This is 
as it should be, (see § 278), remembering that the subtan- 
gent of a parabola is bisected by the vertex. 

282. I-Beam of Uniform Strength. — Support and load same 
as in the preceding §. Fig. 288. Let the area of the 



338 



MECHANICS OF ENGINEERING. 




Fig. 288. 

flange -sections be = F and let it be the same for all values 
of x. Considering all points of F at any one section as at 
the same distance z from the neutral axis, we may write 
/ = z 2 F, and assuming that the flanges take all the tension 
and compression while the (thin) web carries the shear, the 
free body of length x in Fig. 288 gives (moms, about n) 

P-=P c x ; i.e. S^- — =P c x : or, since p is to be constant, 

e z 

z = [Const.], x (1) 

i. e. z must be made proportional to x. 

Hence the flanges should be made straight Practically, 
if they unite at C, the web takes but little shear. 



283. Rectang. Section. Height Constant. Two Supports (at Ex- 
tremities). Single Eccentric Load. . p 
—Fig. 289. b and h are the f c p 

B. With BO free we have 
P^Is-P l =0.:p B = 6 J^ (1) 



W 




Fig. 289. 



At any other section on BO, as n, where the width is % 
the variable whose relation to x is required, we have for 
n free 



6P ti x 



P^=P x;or^l=P x., Pa 
e n y 2 h uh 2 

Equating p B and p n we have u :b :: x :l 



(2) 
(3) 



FLEXURE. BEAMS OF UNIFORM STRENGTH. 



339 



That is, BO must be wedge-shaped with its edge at 0, ver- 
tical. 

284. Similar Rectang. Sections. Otherwise as Before. — Fig. 289 a, 
b and h are the dimen- 
sions at B; at any other 
section n y on BO, the height 
v and width u, are the _° 
variables whose relation to 
x is desired and by hypoth- 
esis are connected by the 
relation 

u \ v v, o ; h . . . . . (1)> 
(since the section at n is a rectangle similar to that at B). 




Fig. 289 a. 



For the free body BO 



Pb = 






For the free body nO p n =^?f 

uv 2 



(2). 
(3). 



Writing p a = p B we obtain l -4- bh 2 = x -r- uv 2 , in which 
put u = bv -4- h, from (1) ; whence 



H ' "WMXVi 



(4) 



which is the equation of the curve (a cubic parabola) 
whose abscissa is x and ordinate j4 v ; i.e., of the upper 
curve of the outline of the central longitudinal vertical 
plane section of the body (dotted line BO) which is sup- 
posed symmetrical about such a plane. Similarly the 
central horizontal plane section will cut out a curve a 
quarter of which (dotted line B' 0) has an equation 



(y 2 uf=(y 2 b?* 



(5) 



That is, the height and width must vary as the cube root 



340 MECHANICS OF ENGINEERING. 

of the distance from the support. The portion CB will 
give corresponding results, referred to the support C. 

[If the beam in this problem is to have circular cross- 
sections, let the student treat it in the same manner.] 

286. Uniform Load. Two End Supports. Rectangular Cr. 

Sections. Width Constant. — j_ wl ^. x ^s. w i 

Weight of beam neglected. * ? _ * I 

How should the height vary, j^fe^L *( "^-^Q^7\ 

the height and width at the j_~f^Tr^ r- } ~ \^' X ~~Z ^7\ 
middle being h and b? Fig. LXT } I | B '[ J [ I [H/ 
289 b. From symmetry each ^ ■ = t| I ' 

reaction = y 2 TV = x / 2 wl. Fig. 289 ». 

At any cross section n, the width is = b, (same as that at 
the middle) and the height = v, variable. I (moms. n ) =0, 
for the free body nO, gives 

^A = y 2ldx - !f ;i e P»/fv 3 = y 2W i x - ™L ... (1) 

e n 2 y 2 v 2 

But for a beam of uniform strength, p a is to be = p B as 
computed from I (moms. B ) = for the free body . . BO, 
i.e from 

P*M®=y 2 wi . l - H'M . . . (2 ) 

y^/i A A 

Hence solve (1) for p n and (2) for p B and equate the results, 

whence v 2 = ^[lx-x 2 ] ; or (y 2 vf=i/^[lx-^] . (3) 

This relation between the abscissa x and the ordinate y 2 v, 
of the curve GBO, shows it to be an ellipse since eq. (3) 
is that of an ellipse referred to its principal diameter and 
the tangent at its vertex as co-ordinate axes. 

In this case eq. (3) covers the whole extent of both 
upper and lower curves, i.e. the complete outline, of the 
curve CBOB', whereas in Figs. 286, 289, and 289 a, such is 
not the case. 



FLEXURE. BODIES OF UNIFORM STRENGTH. 341 

287. Cantilevers of Uniform Strength. — Beams built in at 
one end, horizontally, and projecting from the wall with- 
out support at the other, should have the forms given be- 
low, for the given cases of loading, if all cross-sections are 
to be Rectangular and the weight of beam neglected. Sides 
of sections horizontal and vertical. Also, the sections are 
symmetrical about the axis of the piece, b and h are the 
dimensions at the wall. 1= length. No proofs given. 




Fig. 290. 




Width constant. -^ 
Vertical outline _. 
parabolic. Single \ Fl §- 290 > (°>- 
end load. 



<^jj 



(y 2 vf=(y 2 hy± (i) 



Height constant. 
Single end load. 
Horizontal outline 
triangular. 



Fig. 290, (b). (y 2 u)=(y 2 b% . (2) 



Constant ratio of -x 
height v to width u. I 
Both outlines cu . f Fig. 290. (c). 

bic parabolas. J 



WM#\fl 



(y 2 uf=(y 2 bf 



(3) 
(3)' 



342 



MECHANICS OF ENGINEERING. 



Uniform Load. 
Width constant. 
Vertical outline tri- 
angular. 

Uniform Load. 
Height constant. 
Horiz. outline is 
two parabolas meet- 
ing at (vertex) 
with geomet. axes 
II to wall. 

Uniform Load.-. 
Both outlines semi- 
cubic parabolas. V 
Sections similar 
rectangles. 



Fig. 291, (a). (y 2 v)=(y 2 h) 



W 



Fig. 291, (b). 



#«=(#% 



(5) 



(y 2 uy=(y 2 by 



(6) 



Fig. 291, (c). 



W=W? (6)' 



289. — Beams and cantilevers of circular cross-sections 
may be dealt with similarly, and the proper longitudinal 
outline given, to constitute them " bodies of uniform 
strength." As a consequence of the possession of this 
property, with loading and mode of support of specified 
character, the following maybe stated; that to find the 
equation of safe loading any cross-section whatever may be 
employed. This refers to tension and compression. As 
regards the shearing stresses in different parts of the beam 
the condition of " uniform strength " is not necessarily ob- 
tained at the same time with that for normal stress in the 
outer fibres. 



DEFLECTION OF BEAMS OF UNIFORM 
STRENGTH. 

290. Case of § 283, the double wedge, but symmetrical, 
i.e., li=l =yi, Fig. 292. Here we shall find the use of the 



BEAMS OF UNIFORM STRENGTH. 



343 




Fig. 292. 



EI 



form — (of the three forms for the moment of the stress 

P 
couple, see eqs. (5), (6) and (7), §§ 229 and 231) of the most 

direct service in determining the form of the elastic curve 
OB, which is symmetrical, and has a common tangent at 
B, "with the curve BC. First to find the radius of curva- 
ture, p, at any section n, we have for the free body nO, 
i'(moms. n =0), whence 

EI /-r, r, i i f from eq. ) u ,_ ,,.,,,. 

— —-+}4Px=0 ; but | ( 3 ) § 283 J x= i# l and /= lv uh 



we have 1 / 1 



E 

9 



uh 3 = 



%P -r and 



?=Vi 



I 



E_ 
P 



(1) 



from which all variables have disappeared in the right 
hand member ; i.e., p is constant, the same at all points of 
the elastic curve, hence the latter is the arc of a circle, 
having a horizontal tangent at B. 

To find the deflection, d, at B, consider Fig. 292, (b) 
where d=KB, and the full circle of radius BIl=p is 
drawn. 

The triangle KOBJs similar to YOB, 
and .-. KB : OB : : OB : YB 
But 0B=y 2 l, KB=d and YB=2p 

■'• 2^T' •"" ' m eq * W' ^ = /*'WE 



(2) 



From eq. (4) §233 we note that for a beam of the same 

material but prismatic (parallelopipedical in this case,) 

having the same dimensions, b and h, at all sections as at 

1 P¥ PI 3 

the middle, deflects an amount =jo 'Wf = /^hh^F r u n ^ e r a 



344 



MECHANICS OF ENGINEERING. 



load P in the middle of the span. Hence the tapering 
beam 01 the present § has only J^ the stiffness of the pris- 
matic beam, for the same &, h, I, E, and P. 

291. Case of §281 (Parabolic Body), With^=? , i.e., Symmet- 
rical. — Fig. 293,(a). Required the equation of the neutral 




Fig. 293. 



line OB. For the free body nO, l(moms. n )=0 gives us 



EI 



d?y 
dxf 



—y 2 Px 



ar 



Fig. 293, (6), shows simply the geometrical relations of the 
problem, position of origin, axes, etc. OnB is the neutral 
line or elastic curv« whose equation, and greatest ordinate 
d, are required. (The right hand member of eq. (1)" is made 
negative because d 2 y-±dx 2 is negative, the curve being con- 
cave to the axis X in this, the first quadrant.) 

Now if the beam were prismatic, 7, the " moment of in- 
ertia " of the cross-section would be constant, i.e., the same 
for all values of x, and we might proceed by taking the x- 
anti-derivative of each member of (1)" and add a constant ; 
but it is variable and is 



=Lbv*=L 

12 12 



bhf 



3_ X 

2 



(#0 

hence (1)" becomes 

12 m 



(fromeq. 3, §281, putting \=y 2 l) 



L #2 



d?y 



dx 2 



Px 



(1)' 



To put this into the form Const, x ^=func. of (x\ we need 



BEAMS OF UNIFORM STRENGTH. 345 

3 

only divide through by x 2 , (and for brevity denote 
1 Ebh\+ (y 2 l)\ by A) and obtain 

We can now take the x -anti -derivative of each member, and 
have 



^=— ^P(2x+^)+ C .... (2)' 

To determine the constant C, we utilize the fact that at B, 
where x=*4l> the slope dy+dx is zero, since the tangent 
line is there horizontal, whence from (2)' 

o=-p v j+c ••■ C=F vT 

.-. (2)' becomes A ^L =P[^/}4l-x #] (2) 

Ct X 

Ay=P[^i.x—y 3 xh+lC'=0] ... (3) 

(C=0 since for x=0, y=0). We may now find the deflec- 
tion d (Fig. 293(6)) by writing x= y 2 l and y=d } whence, after 
restoring the value of the constant A, 

p73 

and is twice as great [being=2. ]* as if the girder 

i 3 

* See § 233, putting I =~ bh in eq. (4). 

were parallelopipedical. In other words, the present girder 
is only half as stiff as the prismatic one. 

292. Special Problem. (I.) The symmetrical beam in Fig. 
294 is of rectangular cross-section and constant width = 5, 



346 



MECHANICS OF ENGINEERING. 



but the height is constant only over the extreme quarter 
spans, being =h L = ^A, i. e., half the height h at mid-span. 
The convergence of the two truncated wedges forming the 
middle quarters of the beam is such that the prolongations 




Fig. 294. 

of the upper and lower surfaces would meet over the supports 
(as should be the case to make A=2^). Neglecting the 
weight of the beam, and placing a single load in middle, it 
is required to find the equation for safe loading ; also the 
equations of the four elastic curves ; and finally the deflec- 
tion. 

The solutions of this and the following problem are left 
to the student, as exercises. Of course the beam here 
given is not one of uniform strength. 

293. Special Problem. (II). Fig. 295. Kequired the man- 
ner in which the width of the beam must vary, the height 
being constant, cross-sections rectangular, weight of beam 




Fig. 295. 



neglected, to be a beam of uniform strength, if the load is 
uniformly distributed ? 



FLEXURE. OBLIQUE FORCES. 347 



CHAPTEK V. 

FLEXURE OF PRISMATIC BEAMS UNDER 
OBLIQUE FORCES. 



294 Remarks, By " oblique forces " will be understood 
external forces not perpendicular to tho beam, but these 
external forces will be confined to one plane, called the 
force-plane, which contains the axis of the beam and also 
cuts the beam symmetrically. The curvature induced by 
these external forces will as before be considered very 
slight, so that distances measured along the beam will be 
treated as unchanged by the flexure. 

It will be remembered that in previous problems the 
proof that the neutral axis of eaeh cross section passes 
through its centre of gravity, rested on the fact that when 
a portion of the beam having a given section as one of its 
bounding surfaces is considered free, the condition of 
equilibrium I (compons. || to beam)=0 does not introduce 
any of the external forces, since these in the problems re- 
ferred to, were "J to the beam ; but in the problems of the 
present chapter such is not the case, and hence the neutral 
axis does not necessarily pass through the centre of gravity 
of any section, and in fact may have only an ideal, geomet- 
rical existence, being sometimes entirely outside of the 
section ; in other words, the fibres whose ends are exposed 
in a given section may all be in tension, (or all in compres- 
sion,) of intensities varying with the distance of each from 
the neutral axis. It is much more convenient, however, to 
take for an axis of moments the gravity axis parallel to the 



348 



MECHANICS OF ENGINEERING. 



neutral axis instead of the neutral axis itself, since this 
gravity axis has always a known position. 



295. Classification of the Elastic Forces. Shear, Thrust, and 
Stress-Couple. Fig. 296. Let AKMbe one extremity of a 
portion, considered free, of a prismatic beam, under oblique 
forces. G is the centre of gravity of the section ex- 
posed, and GG the gravity axis "J to the force plane GAK. 
The stresses acting on the elements of area (each =dF) of 
the section consist of shears (whose sum=e7, the "total 
shear") in the plane of the section and parallel to the force 
plane, and of normal stress parallel to AK and proportion- 
al per unit of area to the distances of the dF's on which 
they act from the neutral axis NC", real or ideal (ideal in 
this figure). Imagine the outermost fibre KA, whose dis- 
tance from the gravity axis is=e and from the neutral axis 



fr-~P* 




Fig. 296. 



=c H-a, to be prolonged an amount AA\ whose length by 
some arbitrary scale represents the normal stress (tension 
oi compression) to which the dF at A is subjected. Then, 
ii a plane be passed through A' and the neutral axis NC" 9 
the lengths, such as mr, parallel to AA' f intercepted between 
this plane and the section itself, represent the stress-inten- 



FLEXURE. OBLIQUE FORCES. 349 

sities (i. e., per unit area) on the respective dF's. (In this 
particular figure these stresses are all of one kind, all ten- 
sion or all compression ; but if the neutral axis occurs 
within the limits of the section, they will be of opposite 
kinds on the two sides of NC") Through C, the point 
determined in A'NG" by the intercept CC of the centre of 
gravity, pass a plane A"M"T" parallel to the section it- 
self ; it will divide the stress -intensity A A' into two parts 
Pi andjo 2 , an( l will enable us to express the stress-intensity 
rar, on any dF at a distance a from the gravity -axis GG, in 
two parts ; one part the same for all dF's, the other depen- 
dent on z, thus, : 

[Stress-intensity on any dF] = Pi+ —p 2 . . (1) 
and hence the 
[actual normal stress on any dF] = p L d F + -^ p 2 dF (2) 

For example, the stress-intensity on the fibre at T, where 
z — — e lt will be p l — — p 2 , and it is now seen Low we may 

G 

find the stress at any dF when p v and p 2 have been found. 
If the distance a, between the neutral and gravity axes is 
desired, we have, by similar triangles 

p 2 : e :: CO: a whence a —^ . e . . . . (3) 

It is now readily seen, graphically, that the stresses or elas- 
tic forces represented by the equal intercepts between the 
parallel planes AMT and A" M" T" , constitute a uniform- 
ly distributed normal stress, which will be called the " uni- 
form thrust," or simply the thrust (or pull, as the case may 

be) of an intensity = p { , and .\ of an amount = Cp x dF = 

p i fdF = p l F. 

It is also evident that the positive intercepts forming the 



350 



MECHANICS OF ENGINEERING. 



wedge A n A'G'G' and the negative intercepts forming the 
wedge M'MG'C' form a system of "graded stresses '* 
whose combination (algebraic) with those of the "thrust " 
shows the two sets of normal stresses to be equivalent to 
the actual system of normal stresses represented by the 
small prisms forming the imaginary solid AMT . . A'MT\ 
It will be shown that these graded stresses constitute a 
" stress-couple." 

Analytically, the object of this classification of the nor- 
mal stresses into a thrust and a stress-couple, may be made 
apparent as follows : 

In dealing with the free body KAM Fig. 296, we shall 
have occasion to sum the components, parallel to the beam, 
of all forces acting (external 
and elastic), also those "1 to 
the beam; and also sum their 
moments about some axis ~J 
to the force plane. Let this 
axis of moments be GG the 
gravity -axis of the section 
(and not the neutral axis) ; 
also take the axis X || to the 
beam and FT to it (and in 
force-plane). Let us see 
what part the elastic forces 
will play in these three summations, 
gives merely a side view. 



p 2 dF-^ 




-ipJF'' 



Fig. 297. 

See Fig. 297, which 
Referring to eq. (2) we see that 



elastic"! 






n a a 



[see eq. (4) § 23]. But as the a's are measured from G a 
gravity axis, z must be zero. Hence 

[The IX of the Elastic forces] =p,F^E j ^ ^^ j (4) 



FLEXURE. OBLIQUE FORCES. 351 

Also, 

[The lYoi the Elastic forces] = J= the shear; . (5) 
while for moments about G [see eq. (1)] 
[The I (moms. G ) of the elastic forces] = 

f\p l dF)z^f( z ^p 2 dF)z 
e i \ 

=p 1 fzdF+SLfz i dF 

and hence finally 

e 

where I Gi — C z 2 dF, is the "moment of inertia " of 
i 
the section about the gravity axis G, (not the neutral axis). 
The expression in (6) may be called the moment of the 
stress-couple, understanding by stress-couple a couple to 
which the graded stresses of Fig. 297 are equivalent. That 
these graded stresses are equivalent to a couple is shown 
by the fact that although they are X forces they do not 
appear in eq. 4, for 2X', hence the sum of the tensions 

Hi. r zdF 1 equals that of the compressions \?1 C zdF 

in that set of normal stresses. 

We have therefore gained these advantages, that, of the 
three quantities J" (lbs.), p v (lbs. per sq. inch), and p 2 (lbs. 
per sq. inch) a knowledge of which, with the form of the 
section, completely determines the stresses in the section, 
equations (4), (5), and (6) contain only one each, and hence 
algebraic elimination is unnecessary for finding any one 
of them ; and that the axis of reference of the moment of 
inertia 7 is the same axis of the section as was used in 
former problems in flexure. 

Another mode of stating eqs. (4), (5) and (6) is this : The 
sum of the components, parallel to the beam, of the exter- 
nal forces is balanced by the thrust or pull ; those perpen- 



352 MECHANICS OF ENGINEERING. 

dicular to the beam are balanced by the shear; while the 
sum of the moments of the external forces about the 
gravity axis of the section is balanced by that of the stress- 
couple. Notice that the thrust can have nojoaoment about 
the gravity axis referred to. 

The Equation for Safe Loading, then, will be this : 



(a) . 0±jp 2 ) max. 

or 

( b ) • (Pi^P-2) max. 



whichever 

is 
greater. 



= £' . . (7) 



For B' ', see table in § 251. The double sign provides for 
the cases where p L and p 2 are oi opposite kinds, one tension 
the other compression. Of course (P1-I-JP2) max is n °t the 
same thing as \_p l max. +p 2 max.]. Inmost cases in prac- 
tice e 1 =e i and then the part (b) of eq. (7) is unnecessary. 

295a. Elastic Curve with Oblique Forces. — (By elastic curve 
is now meant the locus of the centres of gravity of the sec- 
tions.) Since the normal stresses in a section differ from 
those occuring under perpendicular forces only in the ad- 
dition of a uniform thrust (or pull), whose effect on the 
short lengths (=dx) of fibres between two consecutive sec- 
tions TJ'V and U V , Fig. 297, is felt equally by all, the loca- 
tion of the centre of curvature B, is not appreciably differ- 
ent from what it would be as determined by the stress - 
couple alone. 

Thus (within the elastic limit), strains being proportional 
to the stresses producing them, if the forces of the stress- 
couple acted alone, the length dx=G G' of a small portion 
of a fibre at the gravity axis would remain unchanged, and 
the lengthening and shortening of the other fibre-lengths 
between the two sections U V and TJ' P, originally parallel, 
would occasion the turning of TJ' V through a small angle 
(relatively to U V ) about G', into the position which it oc- 
cupies in the figure (297), and G B l would be the radius of 
curvature. But the effect of the uniform pull (added to 
that of the couple) is to shift TJ' V parallel to itself into 
the position UV, and hence the radius of curvature of the 



.FLEXURE. OBLIQUE FORCES. 



353 



elastic curve, of which G G is an element, is G B instead 
of G B'. But the difference between G B and G Q B' is very 
small, being the same, relatively, as the difference between 
G Q G and G G' ; for instance, with wrought-iron, even U p l9 
the intensity of the uniform pull, were as high as 22,000 
lbs. per sq. in. [see § 203] G G would exceed G G' by only 
y i2 of one per cent. (=0.0008) ; hence by using GB' instead 
of GB as the radius of curvature p, an error is introduced 
of so small an amount as to be neglected. 

But from § 231, eqs. (6) and (7), — 

P 
the sum of the moments of the external forces ; hence for 

prismatic beams under oblique forces we may still use 



dxr 



±M% { =M) 



(1) 



as one form for the -(moms.) of the elastic forces of the 
section about the gravity-axis ; remembering that the axis 
X must be taken parallel to beam. 



296. Oblique Cantilever with Terminal Load. — Fig. 298. Let 
I = length. The " fixing " of the lower end of the beam is 
its only support. Measure x along the beam from 0. Let 




Fig. 298. 



Fig. 299. 



n be the gravity axis of any section and nT, =x sin a, the 
length of the perpendicular let fall from n on the line of 
action of the force P (load). The flexure is so slight that 
nT is considered to be the same as before the load is al- 



354 MECHANICS OF ENGINEERING. 

lowed to act. [If a were very small, however, it is evident 
that this assumption would be inadmissible, since then a 
large proportion of nT would be due to the flexure caused 
by the load.] 

Consider nO free, Fig. 299. In accordance with the pre- 
ceding paragraph (see eqs. (4), (5), and (6)) the elastic 
forces of the section consist of a shear J, whose value may 
be obtained by writing 1 7=0 

whence J=P sin a ; . . . . (1) 

of a uniform thrust =piF, obtained from 2X=0, viz : 

P cos a—p l F=0 .'. p 1 F=P cos a ; . (2) 

and of a stress-couple whose moment [which we may write 

either ^-, or EI -M. 1 is determined from, J(moms. u )=0 or 
e axr 

PJ—Px sin a=0, or ^=Px sin a . . (3) 

e e 

As to the strength of the beam, we note that the stress-in- 
tensity, p lt of the thrust is the same in all sections, from 
to L (Fig. 298), and that p 2 , the stress-intensity in the outer 
fibre, (and this is compression if e=no' of Fig. 299) due to 
the stress-couple is proportional to x ; hence the max. of 
LP1+P2] will be in the lower outer fibre at L, Fig. 298, 
where x is as great as possible, =1 ; and will be a compres- 
sion, viz. : 

fo+ft] max.=P rcos g+ ;(si n q)e -| _ _ (4) 

.*. the equation for Safe Loading is 

N/ = pf cosa + Z(sin«)e ~| ^ . _ ^ 

since with e^e, as will be assumed here,[p — — p 2 ] max. 



FLEXURE. OBLIQUE FORCES. 



355 



can not exceed, numerically, [pi+p 2 ~\ max. The stress- 
intensity in the outer fibres along the upper edge of the 
beam, being =p v — p 2 (supposing e l =e) will be compressive 
at the upper end near 0, since there p 2 is small, x being 
small ; but lower down as x grows larger, p 2 increasing, a 
section may be found (before reaching the point L) where 
p 2 =Pi and where consequently the stress in the outer fibre 
is zero, or in other words the neutral axis of that section 
passes through the outer fibre. In any section above that 
section the neutral axis is imaginary, i.e., is altogether out- 
side the section, while below it, it is within the section, but 
cannot pass beyond the gravity axis. Thus in Fig. 300, O'U 




Fig. 300. 



Fig. 301. 



is the locus of the positions of the neutral axis for successive 
sections, while OL the axis of the beam is the locus of the 
gravity axes (or rather of the centres of gravity) of the 
sections, this latter line forming the " elastic curve " un- 
der flexure. As already stated, however, the flexure is to 
be but slight, and a must not be very small. For in- 
stance, if the deflection of from its position before flex- 
ure is of such an amount as to cause the lever-arm OR of 
P about L to be greater by 10 per cent, than its value 
(=1 sin a) before flexure, the value of p 2 as computed from 
eq. (3) (with x=l) will be less than its true value in the 
same proportion. 

The deflection of from the tangent at L, by § 237, Fig. 
229(a) is d=(P sin a)P -^3 EI, approximately, putting P sin a 



356 



MECHANICS OF ENGINEERING. 



for the P of Fig. 229 ; but this very deflection gives to the 
other component, P cos #, || to the tangent at Z, a lever 
arm, and consequent moment, about the gravity axes of all 
the sections, whence for I (moms. L )=0 we have, (more ex- 
actly than from eq. (3) when x=T) 



M= P ( Sin a)l+ PooaaSI^E 



(6) 



(We have supposed P replaced by its components || and 
~| to the fixed tangent at L, see Fig. 301). But even (6) 
will not give an exact value for p 2 at L ; for the lever arm 
of P cos «, viz. d, is >(P sina)? 3 -j-3-ET, on account of the 
presence and leverage of P cos a itself. The true value of 
d in this case may be obtained by a method similar to that 
indicated in the next paragraph. 

297. Elastic Curve of Oblique Cantilever with Terminal Load. 
More Exact Solution. For variety place the cantilever as in 

Fig. 302, so that the deflection 
OY=d tends to decrease the 
moment of P about the gravity 
axis of any section, n. We 
may replace P by its X and Y 
components, Fig. 303, || and 
~| respectively to the fixed 
tangent line at L. The origin, 
0, is taken at the free end of 
-the beam. Let «= angle bet- 




Fig. 303. 



Fig. 302. 

ween P and X. For a free body On, n being any section, 
we have 2 (moms. n )=0 

whence EI d * y 



dx* 



= P(cos a)y — P(sin a) x 



(i) 



[See eq. (1) § 295a]. In this equation the right hand 
member is evidently (see fig. 303) a negative quantity; 
this is as it should be, for EId 2 y-7-dx 2 is negative, the curve 
being concave to the axis X in the first quadrant. (It 
must be noted that the axis X is always to be taken || to 
the beam, for Eldty^-dx 2 to represent the moment of the 
stress-couple.) 



FLEXURE. OBLIQUE FORCES. 357 

Eq. (1) is not in proper form for taking the a>anti-deri- 
vative of both members, since one term contains the vari- 
able y, an unknown function of x. Its integration is in- 
cluded in a more general case given in some works on cal- 
culus, but a special solution by Prof. Kobinson, of Ohio, 
is here subjoined for present needs. * 

"We thus obtain as the equation of the elastic curve in 
Fig. 303, 



/Pcosa r e « : +e^ l T(sina)x— ^cos%"|=sinare^— -e^ x l. (2) 

In which e a denotes the Naperian Base= 2.71828, an abstract 
number, and q for brevity stands for VFcosa-^UI. 

To find the deflection d. we make x=l in (2), and solve 
for y; the result is d. 

The uniform thrust at L is piF=Pcosa .... (3) 
while the stress intensity p 2 in the outer fibre at X, is ob- 

* Denoting Pcoa-t-EIhy q* and Psin a-^-JETby p*, eq. (1) becomes--^ —q^y—p^x . . (6) 

Differentiate (6) then ~^-|=0 2 — f- ~P*- Differentiate again : whence d ^ r =q*— -|- . (7) 
dx ax dx dx 

Letting %=u . (S^so that u=q*y-p*x, from (6) J we ^veg-=-^-and ^=^ 2: 
See (7) -^i =q i u ^ W hich, mult, by 2 die, gives -—^ 2 du du=2q*udu 

• '• ( <ta constant J, ^-j / 2 die du=2q* J udu+C •'• ^=? 2w M-C. whence- 
^q-fV^h^ 1 ^ X= ^ e [ W+ V^ ]~| ftvC, or, putting 

. . e = Nap. base, /*=-4 [u+ V^+C 1 ; or C'e -u= Ju^+C . . (9), 
n n c ' L £ 2 J n «? a 

— oj 

a 2g-:c oa; /-r qx Ce„ 

"e —2C'e u+u*=u*+^-, .'. u=% C'e — — 5_ 
n n <? n 2C'q* 



—qx ) 
qx Ce„ (which Rives y ., ,., .. ., 
A (see eqs. 6 and 8) q*y-p*x=% C'e n ——^- fas tunc. x. Consolidating the 

2 C'q 2 ) 



358 MECHANICS OF ENGINEERING. 

tained from the moment equation for the free body L 
viz: ^=P(sin«)Z— P(cos«)d (4) 

in which e= distance of outer fibre from the gravity axis. 
The equation for safe loading is written out by placing 
the values of p l9 p 2i and d, as derived from equations (2), 
•(3), and (4) in the expression 

To solve the resulting equation for P, in case that is the 
unknown quantity, can only be accomplished by successive 
assumptions and approximations, since it occurs trans- 
cendent ally. 

In case a horizontal tension-member of a bridge-truss is 
subjected to a longitudinal tension P' (due to its position 
in the truss and the load upon the latter) and at the same 
time receives a vertical pressure =P" at the middle, each 
half will be bent in the same manner as the cantilever in 
Fig. 302 ; y 2 P" corresponding to P sin «, and P' to Pcos a. 



p-i qx ~qx 

constant factors we now write y = -^,x-\-me —n« • • the equation required . (10) 



To determine the constants m andn(m— C'+2q 2 ; n= C-*-2C'q 4 ) we first find dy+dx, i.e. 

dv v" 2 $ x — Q x 

by differentiating (10) -^r-~ 2 +y me n +fl^« n .... (11) z=0fory=0 



.'. (10) gives . . . 0—0-{-me n — ne Q =0 i.e. m— n=0 .'. m=n . (12) Also for x=l 

dv p 2 & —of\ 

-^-=0.-. (11) gives 0=~f q ^ n + ne n • • (13); .-.with n=m we have m=n= 

r & -^i 

■i-q 3 « n -H n • • -(14). The equation of the curve, then, substituting (14) in 



-P 



qx —qx 



(10) isy=^x—P- -L— -?__ . (15) .-., Substituting f or p and q we 



have. 



/Pcosa T & —Q l ~\ I r Q x —Q x 1 

(as in §297) ' EI e n + e n xsina—ycosa =sin« e —e n 



FLEXURE. OBLIQUE FORCES. 



359 



298. Inclined Beam with Hinge at One End. — Fig. 304 Let 
«e = e x . Required the equation for safe loading ; also the 
maximum shear, there being but one load, P, and that in 
the middle, The vertical wall being smooth, its reaction, 




H, at is horizontal, while that of the hinge-pin being un- 
known, both in amount and direction, is best replaced by 
its horizontal and vertical components B and V , unknown 
in amount only. Supposing the flexure slight, we find 
these external forces in the same manner as in Prob. 1 § 
37, by considering the whole beam free, and obtain 



H=~ cota ; H also = £. cot« ; V 



a) 



For any section n between and B, we have, from the 
free body nO, Fig. 305, 



uniform thrust = p Y F =5cosa 
and from I (moms. n ) = 0, 

±— - =Hx sin a 

e 



■ (2) 



(3) 



and the shear = J = Hsin a = }4 P cos a (4) 

The max. (Pi+p 2 ) to be found on OB is .*. close above B, 
where x = % I, and is 



360 MECHANICS OF ENGINEERING. 

H cos a , Hie sin a -, . -, „ rcot a , le 



F 



esma i . , „ rcot a . ten /ir s 

-— which = P cos a\ +_ (o) 

21 I 2F 4/J V ' 



In examining sections on CB let the free body be Cn r , 
Fig. 306. Then from I (longitud. comps.) = 

(the thrust=) p Y F —V^ sin a + B cos a (6)' 

i.e. p l F=P[sin a + ]/ 2 cos « cot a] (6); 

while, from 2'(moms. n ') = 0, 

P^= V x' cos a—H x' sin « (7)' 

e 

i.e. ^i=i^Pcosax' (7) 

Hence (p y + ^ 2 ) for sections on CB is greatest when x r 
is greatest, which is when x' = y 2 I, x' being limited be- 
tween x' = and x' = y 2 I, and is 

(p 1+ft ) max. on CB=P cos «[ tan a +p cot »+^ -jf] (8) 

which is evidently greater than the max. (Pi+p 2 ) on BO; 
see eq. (5). Hence the equation for safe loading is 

D , D rtana + ^cota, T/ Ze~| /m 

i2' = P cos «| X^ +%j\ .... (9)' 

in which R' is the safe normal stress, per square unit, for 
the material. 

The shear, J, anywhere on 0^,fromJT (transverse comp.) 
=0 in Fig. 306, is 

J= Vq cos a — H Q sin a = j4 P cos a , . (10) 



361 



FLEXURE. OBLIQUE FORCES. 



As showing graphically all the results found, moment, 
thrust, and shear diagrams are drawn 
in Fig. 307, and also a diagram whose 
ordinates represent the variation of 
(pi+p 2 ) along the beam. Each ordi- 
nate is placed vertically under the 
gravity axis of the section to which it 
refers. 

299. Numerical Example of the Forego- , NCH . 
ing. — Fig. 308. Let the beam be of 
wrought iron, the load P = 1,800 lbs., 
hanging from the middle. Cross sec- 
tion rectangular 2 in. by 1 in., the 2 
in. being parallel to the force-plane. 
Required the max. normal stress in 
any outer fibre ; also the max. total 
shear. , 

This max. stress -intensity will be in 
the outer fibres in the section just below B and on the 
upper side, according to § 298, and is given by eq. (8) of 
that article ; in which, see Fig. 308, we must substitute 
(inch -pound-second-system) P = 1,800 lbs.; F = 2 sq. in.; 
I = V120 2 + 12 2 = 120.6 in.; e = 1 in., I = y 2 W = r 2 = % 
biquad. inches ; cot a = JL ; cos a = .0996 ; and tan a — 10. 




FiG. 307. 



.-. max.( j p 1 +jp 2 )=1800x.0996 



lo+y, 



120.6 



- 1 ]- 




Fig. 308. 



Fig. 309. 



Fig. 310. 



9000 lbs. per sq. inch, very nearly, compression. This i 



is m 



362 MECHANICS OF ENGINEERING. 

the upper outer fibre close under B. In the lower outer fibre 
just under B we haye a tension = p 2 ~Pi — 7,200 lbs. per 
sq. in. (It is here supposed that the beam is secure against 
yielding sideways.) 

300. Strength of Hooks. — An ordinary hook, see Fig. 309, 
may be treated as follows : The load being = P, if we 
make a horizontal section at AB, whose gravity axis g is 
the one, of all sections, furthest removed from the line of 
action of P, and consider the portion C free, we have the 
shear = J = zero (1) 

the uniform pull = p^F—P . . . (2) 

while the moment of the stress-couple, from 2 (moms. g ) = 
0,is 

?*I=Pa (3) 

e 

For safe loading p i + p. 2 must = B', i.e. 

b ' =p [^+t] • • • • (4) 

Tt is here assumed that e = e l9 and that the maximum 
\_P1-\-p2] occurs at AB. 

301 Crane. — As an exercise let the student investigate the 
strength of a crane, such as is shown in Fig. 310. 



FLEXUKE. LONG COLUMNS. 363 



CHAPTER VI. 



FLEXURE OF " LONG COLUMNS." 



302. Definitions. — By " long column " is meant a straight 
beam, usually prismatic, which is acted on by two com- 
pressive forces, one at each extremity, and whose length 
is so great compared with its diameter that it gives way 
(or " fails ") by buckling sideways, i.e. by flexure, instead 
of by crushing or splitting like a short block (see § 200). 
The pillars or columns used in buildings, the compression 
members of bridge-trusses and roofs, the " bents " of a 
trestle work, and the piston-rods and connecting-rods of 
steam-engines, are the principal practical examples of long 
columns. That they should be weaker than short blocks 
of the same material and cross-section is quite evident, but 
their theoretical treatment is much less satisfactory than 
in other cases of flexure, experiment being very largely 
relied on not only to determine the physical constants 
which theory introduces in the formulae referring to them, 
but even to modify the algebraic form of those formulae, 
thus rendering them to a certain extent empirical. 

303. End Conditions. — The strength of a column is largely 
dependent on whether the ends are free to turn, or are 
fixed and thus incapable of turning. The former condi- 
tion is attained by rounding the ends, or providing them 
with hinges or ball-and-socket-joints ; the latter by facing 
off each end to an accurate plane surface, the bearing on 
which it rests being plane also, and incapable of turning. 



304 



MECHANICS OF ENGINEERING. 



In the former condition the column is spoken of as having 
round ends ; Fig. 311, (a) ; in the latter as having fixed ends„ 
(or flat bases ; or square ends), Fig. 311, (b). 




{a) 

\V IK/ 

Fig. 311. Fig. 312. 

Sometimes a column is fixed at one end while the other 
end is not only round but incapable of lateral deviation from 
the tangent line of the other extremity ; this state of end 
conditions is often spoken of as "Pin and Square," Fig. 
311, (c). 

If the rounding of the ends is produced by a hinge or 
" pin joint," Fig. 312, both pins lying in the same plane 
and having immovable bearings at their extremities, the 
column is to be considered as round-ended as regards flex- 
ure in the plane ~| to the pins, but as square-ended as re- 
gards flexure in the plane containing the axes of the pins. 

The " moment of inertia " of the section of a column will 
be understood to be referred to a gravity axis of the sec- 
tion which is T to the plane of flexure (and this corres- 
ponds to the " force-plane " spoken of in previous chap- 
ters), or plane of the axis of column when bent. 

303a. Euler's Formula. — Taking the case of a round-ended 
column, Fig. 313 (a), assume the middle of the length as 
an origin, with the axis X tangent to the elastic curve at 
that point. The flexure being slight, we may use the form 
EI tfy-i-dx* for the moment of the stress-couple in any 



FLEXUEE. LONG COLUMNS. 



365 






dy 


dy 


dy 


dx- 


— V 


r-^-p 


1^6 


dx- 


H/ 






dx 








~V~ 


fo~ 


-a-y. 


< 


Js 

T i 


_a— 


— , 




\x 

1 

1 

1 

1 




Y 



Fig. 313. 



Fig. 314. 



section n, remembering that with this notation the axis X 
must be || to the beam, as in the figure (313). Considering 
the free body nC, Fig. 313 (b), we note that the shear is 
zero, that the uniform thrust =P, and that 2 , (moms. n )=0 
gives (a being the deflection at 0) 

Multiplying each side by dy we have 
El 



(1) 



dx' 



dy d?y—Pa dy — Py dy 



(2) 



Since this equation is true for the y, dx, dy, and d 2 y of any 
element of arc of the elastic curve, we may suppose it 
written out for each element from where y—0, a,nddy=0 J 
up to any element, (where dy—dy and y =y) (see Fig. 314) 
and then write the sum of the left hand members equal to 
that of the right hand members, remembering that, since 
dx is assumed constant, 1-^-dx 2 is a common factor on the 
left. In other words, integrate between and any point 
of the curve, n. That is, 



dy=dy 



ay=ay 

/ [dy]d[dy] =Pa £ d y~ P £ V d V ( 3 ) 
The product dy d 2 y has been written (dy)d(d,y), (for d 2 y is 



dx 2 



366 MECHANICS OF ENGINEERING. 

the differential or increment of dy) and is of a form like 
xdx, or ydy. Performing the integration we have 
EI dy 2 j> ^y 2 

which is in a form applicable to any point of the curve, 
and contains the variables x and y and their increments 
dx and dy. In order to separate the variables, solve for dx, 
and we have 



PV2ay-y* V j> , /y . . (5) 



dx=jm dy otdx = IEI 



i.e.,cc=±y-p (vers, sin - X ^J . . . (6) 

(6) is the equation of the elastic curve BOG, Fig. 313 (a), 
and contains the deflection a. If P and a are both given, 
y can be computed for a given #, and vice versa, and thus 
the curve traced out, but we would naturally suppose a to 
depend on P, for in eq. (6) when x— y 2 l, y should =a. Mak- 
ing these substitutions we obtain 

#*= V? < vers - sin ~' 10 °) ; ie - # ,= yl§ I (7) 

Since a has vanished from eq. (7) the value for P ob- 
tained from this equation, viz.: 

r„=m* .... (8) 

is independent of a, and 

is „\ to be regarded as that force (at each end of the round- 
ended column in Fig. 313) which will hold the column at 
any small deflection at which it may previously have been 



FLEXUEE. LOXG COLUMNS. 



3G7 



In other words, if the force is less than P no flexure at 
all will be produced, and hence P is sometimes called the 
force producing " incipient flexure." [This is roughly ver- 
ified by exerting a downward pressure with the hand on 
the upper end of the flexible rod (a T-square-blade for in- 
stance) placed vertically on the floor of a room ; the pres- 
sure must reach a definite value before a decided buckling 
takes place, and then a very slight increase of pressure oc- 
casions a large increase of deflection.] 

It is also evident that a force slightly greater than P 
would very largely increase the deflection, thus gaining for 
itself so great a- lever arm about the middle section as to 
cause rupture. For this reason eq. (8) may be looked 
upon as giving the Breaking Load of a column with round 
ends, and is called Euler' s formula. 

Keferring now to Fig. 311, it will be seen that if the three 
parts into which the flat-ended column is di- 
vided by its two points of inflection A and B 
are considered free, individually, in Fig. 315, 
the forces acting will be as there shown, viz.: 
At the points of inflection there is no stress- 
couple, and no shear, but only a thrust, =P> 
and hence the portion AB is in the condition 
of a round-ended column. Also, the tangents 
to the elastic curves at and C being pre- 
served vertical by the f rictionless guide-blocks 
and guides (which are introduced here simply 
as a theoretical method of preventing the ends 
from turning, but do not interfere with verti- 
cal freedom) OA is in the same state of flex- 
ure as half of AB and under the same forces. 
Hence the length AB must = one half the 
total length I of the flat-ended column. In 
other words, the breaking load of a round- 
ended column of length =y 2 l, is the same as 
that of a flat-ended column of length =1. 
Hence for the I of eq. (8) write y 2 l and we 
have as the breaking load of a column with 
flat-ends and of length =?. 




y,i 



Fig. 315. 



308 MECHANICS OF ENGINEERING. 

Pi=±EI* . . . . (9) 

Similar reasoning, applied to the " pin-and-square " 
mode of support (in Fig. 311) where the points of inflec- 
tion are at B, approximately % I from (7, and at the 
extremity itself, calls for the substitution of 2 /?> I for I in 
eq. (8), and hence the breaking load of a " pin-and-square " 
column, of length = I, is 

P '=I m i • •, • ( 10 ) 

Comparing eqs. (8), (9), and (10), and calling the value of 
P x (flat-ends) unity, we derive the following statement : 
The breaking loads of a given column are as the numbers 



1 

flat-ends 



9/16 
pin-and-square 



74 
round-ends 



according to the 
mode of support. 



These ratios are approximately verified in practice. 

Euler's Formula [i.e., eq. (8) and those derived from it, 
(9) and (10)] when considered as giving the breaking load 
is peculiar in this respect, that it contains no reference to 
the stress per unit of area necessary to rupture the material 
of the column, but merely assumes that the load producing 
" incipient flexure ", i.e., which produces any bending at 
all, will eventually break the beam because of the greater 
and greater lever arm thus gained for itself. In the canti- 
lever of Fig. 241 the bending of the beam does not sensibly 
affect the lever-arm of the load about the wall-section, but 
with a column, the lever- arm of the load about the mid- 
section is almost entirely due to the deflection produced. 

304. Example. Euler's formula is only approximately 
verified by experiment. As an example of its use when 
considered as giving the force producing " incipient flex- 
ure " it will now be applied in the case of a steel T-square- 
blade whose ends are free to turn. Hence we use the 
round-end formula eq. (8) of §303, with the modulus of 
elasticity 27=30,000,000 lbs. per sq. inch. The dimensions 



FLEXURE. LOXG COLUMNS. 369 

are as follows : the length I = 30 in., thickness = ^ of an 
inch, and width = 2 inches. The moment of inertia, 7, 
about a gravity axis of the section || to the width (the 
plane of bending being || to the thickness) is (§247) 



/ 1 \ 3 1 

X 2 xf 3qJ = jgg-QQgbiquad. inches. 



7 1 *» 1 ~ /lx3 
1 = 12 M -12 



..-. , with tz = 22 -7- 7, 

7T 30,000,000 22 2 1 _ Qrv> „ 
Po-^/i p - 162,000 * 7 2 ' 900- 2 * 031bs - 

Experiment showed that the force, a very small addition 
to which caused' a large increase of deflection or side-buck- 
ling, was about 2 lbs. 

305. Hodgkinson's Formulae for Columns. — The principal 
practical use of Euler's formula was to furnish a general 
form of expression for breaking load, to Eaton Hodgkin- 
son, who experimented in England in 1840 upon columns 
of iron and timber. 

According to Euler's formula we have for cylindrical 

columns, /being = % izr^ = JL iz& (§247), 

for flat-ends . . P x = * Et? . ^ 

i.e., proportional to the fourth power of the diameter, and 
inversely as the square of the length. But Hodgkinson's 
experiments gave for wrought-iron cylinders 

^3.55 ,73.5 

P x = (const.) x 72 — ; and for cast iron P x = (const.) x Tl - 

Again, for a square column, whose side = I, Euler's for- 
mula would give 

^while Hodgkinson found for square pillars of wood 

P^constOx^ 



370 MECHANICS OF ENGINEERING. 

Hence in the case of wood these experiments indicated the 
same powers for b and I as Euler's formula, but with a dif- 
ferent constant factor ; while for cast and wrought iron 
the powers differ slightly from those of Euler. 

Hodgkinson's formulae are as follows, and evidently 
not homogeneous ; the prescribed units should .*. be care- 
fully followed, d denotes the diameter of the cylindrical 
columns, b the side of square columns, 1= length. 

("For solid cylindrical cast iron columns, flat-ends ; 

< Breaking load in tons ) A A -, n , -, . . -, x 3 - 56 /7 . ,. \ 7 
\ of 2,240 lbs. each [ =44.16 x (d m inches) +.Q m ft.) 

( For solid cylindrical wrought iron columns, flat-ends ; 

< Breaking load in tons ) 10 , , , . . , x 3 - 55 /7 . ., J 
\ of 2,240 lbs. each } =134 X (dm inches) - Q in ft.) 

f For solid square columns of dry oak, fiat-ends ; 
\ Breaking load in tons ) 1An . ,, . . , ^ fl . ,, , 2 
I of 2,240 lbs. each | = 10 - 95 X ( 6 m mclles ) + lnfi ) 

["For solid square columns of dry fir, fiat-ends ; 

Hodgkinson found that when the mode of support was 
" pin-and-square," the breaking load was about y 2 as 
great ; and when the ends were rounded, about ^ as great 
as with flat ends. These ratios differ somewhat from the 
theoretical ones mentioned in §303, just after eq. (10.) 

Experiment shows that, strictly speaking, pin ends are 
not equivalent to round ends, but furnish additional 
strength ; for the friction of the pins in their bearings 
hinders the turning of the ends somewhat. As the lengths 
become smaller the value of the breaking load in Hodg- 
kinson's formulae increases rapidly, until it becomes larger 
than would be obtained by using the formula for the 
crushing resistance of a short block (§201) viz., FC, i.e., 
the sectional area X the crushing resistance per unit of 
area. 

In such a case the pillar is called a short column, or " short 
block," and the value FC is to be taken as the breaking 



FLEXURE. LOXG COLUMNS. 371 

load. This distinction is necessary in using Hodgkinson's. 
formulae ; i.e., the breaking load is the smaller of the two 
values, FC and that obtained by Hodgkinson's rule. 

In present practice Hodgkinson's formulae are not often 
used except for hollow cylindrical iron columns, for which 
with d 2 and d v as the external and internal diameters, we 
have for flat-ends 

Breaking load in tons ) _ n , (d 2 in in.) 355 — (d x in in.) 3 - 55 
of 2,240 lbs. each } -^ onst - X (« in feet)" " 

in which the const. = 44.16 for cast iron, and 134 for 
wrought, while n = 1.7 for cast-iron and = 2 for wrought. 

306. Examples of Hodgkinson's Formulae. — Example 1. Ee- 
quired the breaking weight of a wrought -iron pipe used 
as a long column, having a length of 12 feet, an internal 
diameter of 3 in., and an external diameter of 3^ inches, 
the ends having well fitted flat bases. 

If we had regard simply to the sectional area of metal, 
which is F = 1.22 sq. inches, and treated the column as a 
short block (or short column) we should have for its com- 
pressive load at the elastic limit (see table §203) P"=FC" 
= 1.22 x 24,000=29,280 lbs. and the safe load P 1 may be 
taken at 16,000 lbs. 

But by the last formula of the preceding article we have 

t^felr 1 \ 1-134.0X1 3 - 25 ) 3 :- 3 " 5 - 15.07 tons 
tons of 2,240 lbs. each J 12 2 

i.e.= 15.07 x 2240=33,768 lbs. 

Detail, [log. 3.25] x 3.55= 0.511883x3.55= 1.817184; 

[log. 3.00] X 3.55=0.477,121x3.55=1.693,779 ; 

and the corresponding numbers are 65.6 and 49.4 ; their 
difference == 16.2, hence 

Br. load in long tons = =15.072 long tons. 

=33,768 lbs. 



372 MECHANICS OF ENGINEERING. 

With a " factor of safety " (see §205) of four, we have, as 
the safe load, P' = 8,442 lbs. This being less than the 
16000 lbs. obtained from the " short block " formula,should 
be adopted. 

If the ends were rounded the safe load would be one- 
third of this i.e., would be 2,814 lbs ; while with pin-and- 
square end-conditions, we should use one-half, or 4,221 lbs. 

Example 2. Kequired the necessary diameter to be 
given a solid cylindrical cast-iron pillar with flat ends, that 
its safe load may be 13,440 lbs. taking 6 as a factor of 
safety. Let d = the unknown diameter. Using the proper 
formula in § 305, and hence expressing the breaking load, 
which is to be six times the given safe load, in long tons 
we have (the length of column being 16 ft.) 



13440 x6_44j.6 (d in inches) 3 - 
2240 ~~" 16 17 



(1) 



i.e. [d in inches] 355 =?|^|i 7 (2) 



or log.d=glj[log. 36+1.7 xlog. 16-log. 44.16] . . ' (3) 
,\ log.d=ig-[1.958278] =0.551627 .-. d = 3.56 ins. 

This result is for flat ends. If the ends were rounded, 
^we should obtain d = 4.85 inches. 

307. Rankine's Formula for Columns. — The formula of this 
name (some times called Gordon's, in some of its forms) has 
a somewhat more rational basis than Euler's, in that it in- 
troduces the maximum normal stress in the outer fibre and 
is applicable to a column or block of any length, but still 
contains assumptions not strictly borne out in theory, thus 
introducing some co-efficients requiring experimental de- 
termination. It may be developed as follows : 

Since in the flat-ended column in Fig. 315 the middle 
portion AB, between the inflection points A and B> is 
acted on at each end by a thrust = P, not accompanied by 
any shear or stress-couple, it will be simpler to treat that 



FLEXURE. LONG COLUMNS. 



37a 



P- 



portion alone Fig. 316, (a), since the thrust and stress- 
couple induced in the section at 
B, the middle of AB, will be equal 
to those at the flat ends, and G 9 
in Fig. 315. Let a denote the de- 
flection of B from the straight line 
AB. Now consider the portion 
AB as a free body in Fig. 316, (b), 
putting in the elastic forces of the 
section at B, which may be clas- 
sified into a uniform thrust = 
PiF, and a stress couple of moment Fig. 316. 





pJ 



(see § 294). (The shear is evidently zero, from 



1' (hor comps.) = 0). Here p { denotes the uniform pres- 
sure (per unit of area), due to the uniform thrust, and p 2 
the pressure or tension (per unit of area), in the elastic 
forces constituting the stress-couple, on the outermost 
element of area, at a distance e from the gravity axis ("] 
to plane of flexure) of the section. F is the total area of 
the section. I is the moment of inertia about the said 
gravity axis, g 

I (vert, comps.) = gives P = p^ . . (1) 
J(moms. g ) =0 gives Pa =^L .... (2) 

For any section, n, between A and B, we would evidently 
have the same p l as at B, but a smaller p 2i since Py < Pa 
while e, I, and F, do not change, the column being pris- 
matic. Hence the max. (pi-\-p 2 ) is on the concave edge at 
B and for safety should be no more than C -f- n, where C 
is the Modulus of Crushing (§ 201) and n is a " factor of 
safety." Solving (1) and (2) for p 1 andp 2 > and putting their 
sum = G -T- n t we have 



P,Pae_ C 
F^ n 



(3) 



We might now solve for P and call it the safe load, but it 



374 'MECHANICS OF ENGINEERING. 

is customary to present the formula in a form for giving 
the breaking load, the factor of safety being applied after- 
ward. Hence we shall make n — 1, and solve for P, call- 
ing it then the breaking load. Now the deflection a is un- 
known, but may be expressed approximately, as follows, 
in terms of e and I. 

Suppose two columns of lengths = V and V\ each 

EI' v 'I' 

bearing its safe load. Then at the point B, — -.._=*!— \ i.e., 

p e 

E'e' = p' p 2 '. Considering the curve AB as a circular arc 
we have (see § 290) a' = V 2 -^- 32 p' , i.e. a' = J^ . -, ; and 

»■/' I" 2 
similarly for the other column, a" = t/L,, . — - . If the 
J 32E" e" 

columns are of the same material i?' = E" , and if each is 
bearing its safe load we may assume p 2 — p 2 " nearly, in 
which case the term p 2 " -*~ E" = Vi ~^~ -^ j an ^ we ma 7 
say that the deflection a, under safe load, is proportional 
to (length) 2 -s- e, approximately, i. e., that ae = pi 2 , where 
/? is a constant (an abstract number also) dependent on 
experiment and different for different materials, and I the 
full length. We may also write, for convenience, /= Fk 2 , 
k being the radius of gyration (see § 85). Hence, finally, 
we have from eq. (3) 

Breaking load ) _ p _ FG /A , 

for flat ends ) 1_ - — 7F .... W 

This is known as Rankine's formula. 

By the same reasoning as in § 303, for a ronnd-ended 
column we substitute 2 I for I ; for a pin-and-square col- 
umn -| I for I ; and .*. obtain 

Breaking load { _ p _ FC ,~\ 

for a round-ended column J ~~ ° W * * ' w 

1+4% 

Breaking load for )_p_ FG ,/»>, 

a pin-and-square column \ ~ 2— W ° ^ ' 

7.2 



ELEXURE. LONG COLUMNS. 



375 



These formulae, (4), (5), and (6), unlike Hodgkinson's, 
are of homogeneous form. Any convenient system of units 
may therefore be used in them. 

Rankine gives the following values for C and J3, to be 
used in these formulae. These are based on Hodgkinson's 
experiments. 





Cast Iron. 


We't Iron. 


Timber. 


C in lbs. per sq. in. 


80,000 


36,000 


7,200 


/3 (abstract number) 


1 

6,400 


1 


1 


36,000 


3,000 



If these numerical values of G are used F must be ex- 
pressed in Sq. Inches and P in Pounds. Rankine recom- 
mends 4 as a factor of safety for iron in quiescent struct- 
ures, 5 under moving loads ; 10 for timber. The N. J. 
Iron & Steel Co. use Rankine's formula for their wrought 
iron rolled beams, when used as columns, with a factor of 
safety of 4j^. 



308. Examples, Using Rankine's Formula. — Example 1. — 
Take the same data for a wrought iron pipe used as a 
column, as in example 1, § 306 ; i.e., 1=12 ft. =144 inches, 
F=%[tt(3}() 2 —7u3 2 ]=1.227 sq. inches, while ¥ for a nar- 
row circular ring like the present section may be put 
= %{iy%f (see § 98) sq. inches. "With these values, and 
(7=36,000 lbs. per sq. in., and ft^^m (for wrought iron), 
we have from eq. (4), for flat ends, 



iY 



1.227x36,000 



1+ 



1 (144) 2 

36,000 ' i [1.625] s 



:30743.6 lbs. 



(1) 



This being the breaking load, the safe load may be taken 
= % or Vs °f 30743.6 lbs., according as the structure of 



376 MECHANICS OF ENGINEERING. 

which the column is a member is quiescent or subject to 
vibration from moving loads. By Hodgkinson's formula 
33,768 lbs. was obtained as a breaking load in this case 
(§ 306). 

For rounded ends we should obtain (eq. 5) 

P o =16,100. lbs., as break, load . (2) 

and for pin-and-square, eq. (6) 

P 2 =24,908. lbs. as break, load . . (3) 

Example 2.— (Same as Example 2, § 306). Required by 

Rankine's formula the necessary diameter, d } to be given 

a solid cylindrical cast-iron pillar, 16 ft. in length, with 

rounded ends, that its safe load may be six long tons (i.e., 

Ttd 2 
of 2,240 lbs. each) taking 6 as a factor of safety. F= — - , 

while the value of h 2 is thus obtained. From § 247, I for 
a full circle about its diameter = %7rr i =7rr 2 .}{r 2 .*. ¥=%r 2 
=l /\S 2 ' Hence eq. (5) of § 307 becomes. 

Po = %**% . . . (1) 

d l 

P the breaking load is to be =6x6x2,240 lbs., for cast- 
iron is 80,000 lbs. per sq. inch, while ft (abstract number) 
— dbb- Solving for d we have the biquadratic equation : 

, 4 _ 28 x6x6x 2,240 , 2 _ 28x6x6x2,240xl6 2 xl2 2 x4 
22x80,000 — " 22x80,000x400 

whence d 2 = 0.641 (1± 33.92), and taking the upper sign, 
finally, d= V22.4 =4.73 inches. (By Hodgkinson's rule 
we obtained 4.85 inches). 

309. Radii of Gyration. — The following table, taken from 
p. 523 of Rankine's Civil Engineering, gives values of &*, 
the square of the least radius of gyration of the given cross- 
section about a gravity-axis. By giving the least value of 



FLEXUKE. LONG COLUMNS. 



377 



^ it is implied that the plane of flexure is not determined 
by the end-conditions of the column ; (i.e., it is implied 
that the column has either flat ends or round ends.) If 
either end (or both) is a pin-joint the column may need to 
be treated as having a flat -end as regards flexure in a plane 
containing the axis of the column and the axis of the pin, 
if the bearings of the pin are firm ; while as regards flex- 
ure in a plane perpendicular to the pin it is to be consid- 
ered round -ended at that extremity. 

In the case of a " thin cell " the value of W is strictly 
true for metal infinitely thin and of uniform thickness ; still, 
if that thickness, does not exceed y& of the exterior diame- 
ter, the form given is sufficiently near for practical pur- 
poses ; similar statements apply to the branching forms. 




■~b- 

U) 



(O id) (o 

Fig. 317. 





Fig. 318. 



Solid Eectangle. 
h= least side. 
Thin Square Cell. 
Side= h. 

Thin Kectangular Cell. 
h= least side. 
Solid Circular Section. 
Diameter =d. 
Thin Circular Cell. 
Exterior diam. = d. 
Angle-Iron of Equal 
ribs 



V=±h> 



h 2 



¥= 



Fig. 317(a). 
Fig. 317 (6). 
Fig. 317(c). 
Fig. 317(d). 
Fig. 317 (e). Jc 2 =^d? 
Fig. 317 (/) ¥=!.¥ 



¥=Ld? 

16 



h+3b 



12 h+b 



378 



MECHANICS OF ENGINEERING, 

Fig. 318 (a). ¥= 



b 2 h 2 



12(6 2 +A 2 ) 
Fig. 318 (b). ¥=!¥ 



¥ 



Fig. 318 (c). ¥= — 



Angle-Iron of unequal 

ribs. 
Cross of equal arms. 
I-Beam as a pillar. 
Let area of web =B. 

" " both flanges 

=A. 

Channel Fig . 3 i 8(d) . F= ^ 2 [_ 

Iron. & v y |_12 (^+#) 4 (^+5) 2 J 

Let area of web =5 ; of flanges =A (both), h extends 
from edge of flange to middle of web. 



12 A+B 



AB I 





Fig. 319. 



PHCENIX COLUMN. 



Fig. 320. 



310. Built Columns. — The " compression members " of 
wrought-iron bridge trusses are generally composed of 
several pieces riveted together, the most common forms 
being the Phoenix column (ring-shaped, in segments,) and 
combinations of channels, plates, and lattice, some of which 
are shown in Figs. 319 and 320. 

Experiments on full size columns of these kinds were 
made by the U. S. Testing Board at the "Watertown Arse- 
nal about 1880. 

The Phoenix columns ranged from 8 in. to 28 feet in 
length, and from 1 to 42 in the value of the ratio of length 
to diameter. The breaking loads were found to be some- 
what in excess of the values computed from Kankine's 
formula ; from 10 to 40 per cent, excess. In the pocket- 
book issued by the Phoenix company they give the follow- 
ing formula for their columns, (wrought-iron.) 



FLEXUKE. LOXG COLUMNS. 379 

Breaking load in lbs. )_ 50,000 F m 

for flat-ended columns j 1+ f ' " ' " * ^ j 

3,000// 

where F = area in sq. in., I = length, and h = external 
diameter, both in the same unit. 

Many different formulae have been proposed by different 
engineers to satisfy these and other recent experiments on 
columns, but all are of the general form of Bankine's. 
For instance Mr. Bouscaren, of the Keystone Bridge Co., 
claims that the strength of Phoenix columns is best given 
by the formula - 

Breaking load in ) _ 38,000 F (e) , 

lbs. for flat-ends, j f W 

1 + 100,000& a 

(F must be in square inches.) 

The moments of inertia, 7, and thence the value of k 2 ~ 
I -~ F, for such sections as those given in Figs. 319 and 
320 may be found by the rules of §§ 85-93, (see also § 258.) 



311. Moment of Inertia of Built Column. Example. — It is pro- 
posed to form a column by joining two I-beams by lattice- 
work, Fig. 321, (a). (While the lattice- work is relied upon 
to cause the beams to act together as one piece, it is not 
regarded in estimating the area F, or the moment of iner- 
tia, of the cross section). It is also required to find the 
proper distance apart = x, Fig. 321, at which these beams 
must be placed, from centre to centre of webs, that the 
liability to flexure shall be equal in all axial planes, i.e. 
that the 1 of the compound section shall be the same 
about all gravity axes. This condition will be ful- 
filled if I Y can be made = l x , (§89), being the centre 
of gravity of the compound section, and X perpendicular 
to the parallel webs of the two equal I-beams. 

Let F' = the sectional area of one of the I-beams, F r 
(see Fig. 321(a) its moment of inertia about its web- axis,// 
that about an axis "J to web. (These quantities can be 



380 



MECHANICS OF ENGINEERING. 



found in the hand-book of the iron company, for each size 
of rolled beam). 
Then the 

total 7 X = 2I' X ; and total 7 Y = 2[> Y + F'( X Y\ 

(see §88 eq. 4.) If these are to be equal, we write them so 
and solve for x, obtaining 



/ 4[Z' X -J' Y ] 

v — ip — 



(1) 



312. Numerically ; suppose each girder to be a 10^ inch 
light I-beam, 105 lbs. per yard, of the N. J. Steel and Iron 
Co., in whose hand-book we find that for this beam I' x = 
185.6 biquad. inches, and I\ = 9.43 biquad. inches, while 
F' = 10.44 sq. inches. With these values in eq. (1) we 
have 



7 4 - ( 


185.6-9.43) 


10.44 


t 


V 


Y 


V 






o 


x 


.. 


< 




L 



^ 



(a) 



~^ 



=M 



7"^ 
o 



S 1 r" 



(ft) 



^ 



S_ 



Fig. 321. 



The square of the radius of gyration will be 

V=2F X +2F'=* 371.2 -^20.88 =17.7 sq. in. . (2) 

and is the same for any gravity axis (see § 89). 

As an additional example, suppose the two I-beams united 
by plates instead of lattice. Let the thickness of the plate 
= t, Fig. 321, (b). Neglect the rivet-holes. The distance 
a is known from the hand-book. The student may derive 
a formula for x> imposing the condition that (total 7 X )= 1^ 



FLEXURE. LONG COLUMNS. 



381 



313. Trussed Girders. — When a horizontal beam is trussed 

W=-2 wl 



jy^u^^^ityi! 




Fig. 323. 

in the manner indicated in Fig. 322, with a single post or 
strut under the middle and two tie-rods, it is subjected to 
a longitudinal compression due to the tension of the tie- 
rods, and hence to a certain extent resists as a column, the 
plane of whose flexure is vertical, (since we shall here sup- 
pose the beam supported later ally. )Taking the case of uni- 
form loading, (total load = W) and supposing the tie-rods 
screwed up (by sleeve nuts) until the top of the post is on 
a level with the piers, we know that the pressure between 
the post and the beam is P' = $/& W (see § 273). Hence 
by the parallelogram of forces (see Fig. 322) the tension 
in each tie-rod is 



<?= 



2 cos 



16 



COS a 



At each pier the horizontal component cf Q is 

P= Q sin a— — JKtan a 
* 16 



• (1) 



Hence we are to consider the half -beam 50 as a " pin-and- 
square " column under a compressive force P= 5 / 16 W tan a 
as well as a portion of a continuous girder over three 
equidistant supports at the same level and bearing a uni- 
form load W. In the outer fibre of the dangerous section, 
0, (see also § 273 and Fig. 278) the compression per sq. 
inch due to both these straining actions must not exceed 
a safe limit, B', (see § 251). In eq. (6) § 307, where P 2 is 
the breaking force for a pin-and-square column, the great- 



382 MECHANICS or ENGINEERING. 

est stress in any outer fibre = C ( = the Modulus of Crush- 
ing) per unit of area. If then we write p col . instead of G 
in that equation, and 5 / 16 IF tan a instead of P 2 we have 

j max. stress due ) = == 5 Wt&n an. , 16 n P ] t 
( to column action j ~ ^ co1 Jg " ^ |_ ' ~9~' "Fj 

while from § 273 eq. (3) we have (remembering that our 
present W represents double the 7^of §273). 

j max. stress due \ _ 1 Wle 1 Wle 

\ to girder action j "~- Pgr ~ 16~~ 7 ~16 ' TF 

By -writing _p coL +p yi =i?'= a safe value of compression per 
unit-area, we have the equation for safe loading 

W[5tana(l+™. ^ + l <i~J=16 F£' . . (2) 

Here I = the half -span 0B f Fig. 322, e = the distance of 
outer fibre from the horizontal gravity axis of the cross 
section, h = the radius of gyration of the section referred 
to the same axis, while F = area of section. ft should be 
taken from the end of §307. 

Example.— If the span is 30 ft. = 360 -in., the girder a 15 
inch heavy I-beam of wrought iron, 200 lbs. to the yard, in 
which e = ^ of 15 = iy 2 inches, .F=20 sq. in., and ¥ = 
35.3 sq. inches (taken from the Trenton Co.'s hand-book), 
required the safe load W, the strut being 5 ft. long. 
From §307, p = 1 : 36,000 ; tan a = 15--5 = 3.00. Hence, 
using the units pound and inch throughout, and putting 
R' = 12,000 lbs. per sq. in. = max. allowable compression 
stress, we have from eq. (2) 

w _ 16x20x12,000 

lsfl-u 16 1 : ( 18Q ) 2 1 1 18Q X 7 ^ =36.6 tons. 
L 1 " 1 " 9 "36,000 ' 35.3 y 35.3 

i. e., 69,111 lbs. besides the weight of the beam. 

If the middle support had been a solid pier, the safe load 
would have been 48 tons ; while if there had been no 
middle support of any kind, the beam would bear safely 



FLEXURE. LONG COLUMNS. 



383 



only 11.5 tons, 
the strut)]. 



[Let the student design the tie-rods (and 



314. Buckling of Web-Plates in Built Girders. — In §257 men- 
tion was made of the fact that very high web plates in 
built beams, such as /beams and box-girders, might need 
to be stiffened by riveting T-irons on the sides of the web. 
(The girders here spoken of are horizontal ones, such as 
might be used for carrying a railroad over a short span of 
20 to 30 feet. 

An approximate method of determining whether such 
stiffening is needed to prevent lateral buckling of the web, 
may be based upon Rankine's formula for a long column 
and will now be given. 

In Fig. 323 we have, free, a portion of a bent I-beam, 
between two vertical sections at a distance apart= li Y = 
the height of the web. In such a beam under forces L *° 
its axis it has been proved (§256) that we may consider 
the web to sustain all the shear, J, at any section, and the 
flanges to take all the tension and compression, which 
form the "stress-couple" of the section. These couples 
and the two shears are shown in Fig. 323, for the two 
exposed sections. There is supposed to be no load on this 
portion of the beam, hence the shears at the two ends are 




Fig. 1)23 



equal. N-ow the shear acting between each flange and the 
horizontal edge of the web is equal in intensity per square 
inch to that in the vertical edge of the web ; hence if the 
web alone, of Fig. 323, is shown as a free body in Fig. 324, 
we must insert two horizontal forces = J", in opposite 



384 MECHANICS OE ENGINEERING. 

directions, on its upper and lower edges. Each of these 
= J since we have taken a horizontal length \ — height 
of web. In this figure, 324, we notice that the effect of 
the acting forces is to lengthen the diagonal BD and 
shorten the diagonal AC, both of those diagonals making 
an angle of 45° with the horizontal. 

Let us now consider this buckling tendency along AC, 
by treating as free the strip AC, of small width = b v This 
is shown in Fig. 325. The only forces acting in the direc- 
tion of its length AC&ve the components along AC of the 
four forces J' at the extremities. We may therefore treat 
the strip as a long column of a length I = hi a/2, of a sec- 
tional area F — bb 1} (where b is the thickness of the web 
plate), with a value of k 2 = l / l2 b 2 (see § 309), and with 
fixed (or flat) ends. Now the sum of the longitudinal 
components of the two J'.'s at A is Q = 2 J ' j4 V2 

= J' V% ; but J' itself — jr. b y 2 b Y a/2, since the small 

rectangle on which J' acts has an area = b y 2 b x a/2, and 
the shearing stress on it has an intensity of (J -5- bh^) per 
unit of area. Hence the longitudinal force at each end of 
this long column is 

Q=\j (1) 

According to eq. (4) and the table in § 307, the safe load 
(factor of safety = 4) for a tvrought-iron column of this 
form, with flat ends, would be (pound and inch) 

p__^i36,000 9,000 56, / x 

Fl ' 1+ i 2^-rrn? ■ • • (2) 

36,000Vi2& 2 T 1,500 b 2 : 

If, then, in any particular locality of the girder (of 
zurought-iron) we find that Q is > P lt i.e. 

if £ is > ^° 00b 2 (pound and inch) . . (3) 
/i\ I hi 



FLEXUliE. LOXG COLUMNS. 385 

then vertical stiff eners will be required laterally. 

"When these are required, they are generally placed at 
intervals equal to h lf (the depth of web), along that part 
of the girder where Q is > P Y . 

Example Fig. 326. — Will stiffening pieces be required 
in a built girder of 20 feet span, bearing a uniform load of 
40 tons, and having a web 24 in. deep and }i in. thick ? 

From § 242 we know that the ^^^^w^tons^^ 

greatest shear, /max., is close to J . — ,..- .-. d 

either pier, and hence we investi- | V — io-— ^, , 
gate that part of the girder first. 

J max. = y 2 W = 20 tons 
=40,000 lbs. 
<.*. (inch and lb.), see (3), 

Fig 326. 

J-_40^000_ 

Z^— 2^- 1666 - 6 ( 4 ) 

while, see (3), (inch and pound), 

9,000x?4 




i-h x ^ 



=905.0 (5) 

1,500 * (157 

which is less than 1666.66. 

Hence stiffening pieces will be needed near the extremi- 
ties of the girder. Also, since the shear for this case of 
loading diminishes uniformly toward zero at the middle 
they will be needed from each end up to a distance of 
ig= of 10 ft. from the middle. 



386 



MECHANICS OF ENGINEERING. 



CHAPTER TIL 



LINEAK ARCHES (OF BLOCKWORK), 



315. A Blockwork Arch is a structure, spanning an opening 
or gap, depending, for stability, upon the resistance to 
cornpresssion of its blocks, or vonssoirs, the material of 
which, such as stone or brick, is not suitable for sustain- 
ing a tensile strain. Above the voussoirs is usually 
placed a load of some character, (e.q. a roadway,) whose 
pressure upon the voussoirs will be considered as vertical, 
only. This condition is not fully realized in practice, 
unless the load is of cut stone, with vertical and horizontal 
joints resting upon voussoirs of corresponding shape (see 
Fig. 327), but sufficiently so to warrant 
its assumption in theory. Symmetry 
of form about a vertical axis will also 
be assumed in the following treatment. 




316. Linear Arches. — For purposes of 
theoretical discussion the voussoirs of 
Fig. 327 may be considered to become 
fig. 32r. infinitely small and infinite in number, 

thus forming a " linear arch," while retaining the same 
shapes, their depth ~| to the face being assumed constant 
that it may not appear in the formulae. The joints 
between them are ~| to the curve of the arch, i.e., adjacent 
voussoirs can exert pressure on each other only in the 
direction of the tangent-line to that curve. 



LINEAK ARCHES. 



387 



317. Inverted Catenary, or Linear Arch Sustaining its Own 
Weight Alone. — Suppose the infinitely small voussoirs to 
have weight, uniformly distributed along the curve, weigh- 
ing q lbs. per running linear unit. The equilibrium of 
such a structure, Fig. 328, is of course unstable but theo- 
retically possible. Kequired the form of the curve when 
equilibrium exists. The conditions of equilibrium are, 
obviously : 1st. The thrust or mutual pressure T between 
any two adjacent voussoirs at any point, A, of the curve 
must be tangent to the curve ; and 2ndly, considering a 
portion BA as a free body, the resultant of H the pres- 




Fig. 328. 



sure at B the crown, and Tat A, must balance R the re- 
sultant of the II vertical forces (i.e.,weights of the elementary 
voussoirs) acting between B and A. 

But the conditions of equilibrium of a flexible, inexten- 
sible and uniformly loaded cord or chain are the very 
same (weights uniform along the curve) the forces being 
reversed in. direction. Fig. 329. Instead of compression 
we have tension, while the || vertical forces act toward in- 
stead of away from, the axis X Hence the curve of equi- 
librium of Fig. 328 is an inverted catenary (see § 48) whose 
equation is 



2/+c=^cp +e "J 



(1) 



See Fig. 330. e = 2.71828 the Naperian Base. The "par- 
ameter " c may be determined by putting x = a, the half 
span, and y= Y, the rise, then solving for c by successive 



388 



MECHANICS OF EXGIXEEKIXG. 



approximations. The " horizontal thrust" or H Qi is = qc, 
while if s = length of arch OA, along the curve, the thrust 
T at any point A is 



T=VH Q 2 +q 2 s 2 (2.) 

From the foregoing it may be inferred that a series of vous- 

soirs of finite dimensions, arranged 

so as to contain the catenary curve, 

with joints "] to that curve and of 

equal weights for equal lengths of 

arc will be in equilibrium, and 

moreover in stable equilibrium on 

account of friction, and the finite 

width of the joints ; see Fig, 331. 




FIG. 331. 



318. Linear Arches under Given Loading. — The linear arches 
to be considered further will be treated as without weight 
themselves but as bearing vertically pressing loads (each 
voussoir its own). 

Problem. — Given the form of the linear arch, itself, it is 
required to find the law of vertical depth of loading under 
which the given linear arch will be in equilibrium. Fig. 
332, given the curve ABC, i.e., the linear arch itself, re- 
quired the form of the curve MON, or upper limit of load- 
ing, such that the linear arch ABC shall be in equilibrium 
under the loads lying between the two curves. The load- 
ing is supposed homogeneous and of constant depth ~\ to 
paper ; so that the ordinates z between the two curves are 
proportional to the load per horizontal linear unit Assume 
a height of load z at the crown, at pleasure ; then required 
the z of any point m as a function of z and the curve 
ABC. 




Fig. 332. 



Fig. 333. 



LINEAR ARCHES. 



389 



Practical Solution. — Since a linear arch under vertical 
pressures is nothing more than the inversion of the curve 
assumed by a cord loaded in the same way, this problem 
might be solved mechanically by experimenting with a 
light cord, Fig. 333, to which are hung other heavy cords, 
or bars of uniform weight per unit length, and at equal 
horizontal distances apart when in equilibrium. By varying 
the lengths of the bars, and their points of attachment, we 
may finally find the curve sought, MON. (See also § 343.) 

Analytical Solution. — Consider the structure in Fig. 334. 
A number of rods of finite length, in the same plane, are in 
equilibrium, bearing the weights P, P l9 etc., at the con- 




Fig. 334 



necting joints, each piece exerting a thrust T against the 
adjacent joint. The joint A, (the " pin " of the hinge), im- 
agined separated from the contiguous rods and hence free, 
is held in equilibrium by the vertical force P (a load) and 
the two thrusts T and T', making angles = and Q' with 
the vertical ; Fig. 335 shows the joint A free. From ^(hor- 
izontal comps.)=0, we have. 

Tamd=T'amd'. 

That is, the horizontal component of the thrust in any rod 
is the same for all ; call it H . .'. 



sin U 



(i) 



390 



MECHANICS OF ENGINEEKING. 



Now draw a line As T to T' and write 2 ( compons. II to 
As)=0 ; whence P sin d'= T sin /?, and [see (1)] 

• P - H » sin P (2) 

sin 6/ sin 6' K } 

Let the rods of Fig. 334 become infinitely small and infi- 
nite in nnmber and the load continuous. The length of 
each rod becomes =ds an element of the linear arch. /3 is 
the angle between two consecutive ds's, 6 is the angle be- 
tween the tangent line and the vertical, while P becomes 
the load resting on a single dx, or horizontal distance be- 
tween the middles of the two ds's. That is, Fig. 336, if 
j= weight of a cubic unit of the 
loading, P=yzdx. (The lamina of 
arch and load considered is unity, 
"| to paper, in thickness.) S =a 
constant = thrust at crown ; 
6=6', and sin fi=ds^-p, (since the 
angle between two consecutive tan- 
gents is = that between two con- 
secutive radii of curvature). Hence 
eq. (2) becomes 



jzdx ■ 



H n ds 
p sin 2 6 



but dx=ds sin 6, 




Fig. 336. 



yz 



H,. 



p sin 3 # 



(3) 



Call the radius of curvature at the crown p , and since 
there z= z and 0=90°, (3) gives yz p =H ; hence (3) may 
be written 



z = 



«o/>0 



p sin 3 



w 



This is the law of vertical depth of loading required. For 
a point of the linear arch where the tangent line is verti- 
cal, sin #=0 and z would = oo ; i.e., the load would be in- 



LINEAR AECHBS. 



391 



Jlnitely high. Hence, in practice, a full semi-circle, for in- 
stance, could not be used as a linear arch. 

319. Circular Arc as Linear Arch. — As an example of the 

preceding problem let us ap- 
ply eq. (4) to a circular arc, 
Fig. 337, as a linear arch. 
Since for a circle p is con- 
stant = r, eq. (4) reduces 
to 




sin 3 # 



(5) 



Fig. 33: 



Hence the depth of loading 
must vary inversely as the cube of the sine of the angle 6 
made by the tangent line (of the linear arch) with the ver- 
tical. 

To find the depth z by construction.— Having z given, 
being the centre of the arch, prolong Qa and make ab = 
z Q ; at b draw a "| to Gb, intersecting the vertical through a 
at same point d ; draw the horizontal dc to meet Ga at 
same point c. Again, draw ce "| to Cc, meeting ad in e ; 
then ae = z required ; a being any point of the linear arch. 
For, from the similar right triangles involved, we have 

z=ab=ad sin d—ac sin 6. sin d=ae sin d sin d sin d 



ae= ° ; i.e., ae—z. Q.E.D. 

S1 ^ [see (5.)] 



320. Parabola as Linear Arch. — To apply eq. 4 § 318 to a 

parabola (axis vertical) as linear arch, we must find values 
of p and p the radii of curvature at any point and the 
crown respectively. That is, in the general formula, 



M 



dy\ 
dx) _ 



dx 2 



we must substitute the forms for the first and second dif- 
ferential co-efficients, derived from the equation of the 



392 



MECHANICS OF ENGINEERING. 





Fig. 339. 

curve (parabola) in Fig. 338, i.e. from x 2 = 2 py ; whence 
we obtain 

-^,or cot d,= — and— -^ =— 
ax p ax 2 p 



P 



sin 3 



(6) 



Hence p =\V_JL _i= p cosec. 3 #, i.e. p 

1-^-p 

At the vertex d — 90° .\ p Q = p. Hence by substituting 
for p and p in eq. (4) of § 318 we obtain 

z=z = constant [Fig. 339] (7) 

for a parabolic linear arch. Therefore the depth of homo- 
geneous loading must be the same at all points as at the 
crown ; i.e., the load is uniformly distributed with respect 
to the horizontal. This result might have been antici- 
pated from the fact that a cord assumes the parabolic 
form when its load (as approximately true for suspension 
bridges) is uniformly distributed horizontally. See § 46 
in Statics and Dynamics. 



321. Linear Arch for a Given Upper Contour of Loading, the 
arch itself being the unknown lower contour. Given the 
upper curve or limit of load and the depth z Q at crown, re- 
quired the form of linear arch which will be in equili- 
brium under the homogenous load between itself and that 
upper curve. In Fig. 340 let MONhe the given upper 
contour of load, z is given or assumed,*;' and z" are the 
respective ordinates of the two curves BA G and MON. 
Required the eqation of BAG 



LINEAR ARCHES. 



393 




Fig. 340. 



Fig 341. 



As before, the loading is homogenous, so that the 
weights of any portions of it are proportional to the 
corresponding areas between the curves. (Unity thick- 
ness "1 to paper.) Now, Fig. 341, regard two consecutive 
ds's of the linear arch as two links or consecutive blocks 
bearing at their junction m the load dP — y (z + z"} dx in 
which y denotes the heaviness of weight of a cubic unit of 
the loading. If T and T are the thrusts exerted on these 
two blocks by their neighbors (here supposed removed) 
we have the three forces dP, T and T' t forming a system 
in equilibrium. Hence from IX =0. 



T cos <p = T coe cp' 



and 



IY=0 gives T sin cp' — T sin <p = dP 



(1) 

(2) 



From (1) it appears that T cos <p is constant at all points 
of the linear arch (just as we found in § 318) and hence 
= the thrust at the crown, = H t whence we may write 

T=H -^ cos <p and T'=H 4- cos cp' . . . (3) 

Substituting from (3) in (2; we obtain 

H (tan <p' — tan <p)=dP (4) 

But tan <p' =^— and tan <p'= — It , (dx constant) 

ax QjX 

while dP = y (z f + *") dx. Hence, putting for convenience 

H — ya 2 , (where a = side of an imaginary square of the 



394 MECHANICS OF ENGINEERING. 

loading, whose thickness = unity and whose weight = H) 
we have. 

£=>' +2 "> •• • ^ 

as a relation holding good for any point of the linear arch 
which is to be in equilibrium under the load included 
between itself and the given curve whose ordinates are z", 
Fig. 340. 

322. Example of Preceding. Upper Contour a Straight Line.— 
Fig. 342. Let the upper contour be a right line and hor- 
izontal ; then the z" of eq. 5 becomes zero at all points of 
ON. Hence drop the accent of z' in eq. (5) and we have 

d 2 z z 
dx 2 d l 

Multiplying which by dz we obtain 

dz f^=±zdz (6) 

dx* a 2 w 

This being true of the z, dz, d 2 z and dx of each element of 
the curve O'B whose equation is desired, conceive it writ- 
ten out for each element between 0' and any point m, and 
put the sum of the left-hand members of these equations 
= to that of the right-hand members, remembering that 
a 2 and dx 2 are the same for each element. This gives 



L f 4A-I fzdz; i.e., * **J:\**\ 
x> I a' 'dx* 2 at 2 2 J 

«y tte=0 %J z=z 

(i) 



aaz _ n \ *o/ .... (7. 



dx= — — = a 

VJ-zl 



<© 



LINEAR AKCHES. 



395 




Fig. 342. 



Fig 343 



Integrating (7.) between 0' and any point m 



[■ 



/, 



i.e., x=a log. t 



-•O-lsV®- 1 ) 

P_l. /r 



^© 



j 



or z- 



i[f+f\ 



(8)' 
(8.) 
(9.) 



This curve is called the transformed catenary since we may 
obtain it from a common catenary by altering all the ordi- 
nates of the latter in a constant ratio, just as an ellipse 
may be obtained from a circle. If in eq. (9) a were = z Q 
the curve would be a common catenary. 

Supposing z and the co-ordinates x Y and z x of the point 
B (abutment) given, we may compute a from eq. 8 by put- 
ting x —x x and z = z { , and solving for a. Then the crown- 
thrust H = ya 2 becomes known, and a can be used in eqs. 
(8) or (9) to plot points in the curve or linear arch. From 
eq. (9) we have 

Xr- X — X -1 1- X — X -l 

(10) 



area 
00' mn 



j- = = J zdx=2j [_e dx+e dx \~ 2 |_ e _e J 



Fig. 343. 

Call this area, A. As for the thrusts at the different 
joints of the linear arch, see Fig. 343, we have crown- 
thrust =H= T a 2 . . . ; . . . (11) 
and at any joint m the thrust 

T=VH 2 +( r Af =rVtf+A 2 .... (12) 



396 MECHANICS OF ENGINEERING. 

323. Remarks. — The foregoing results may be utilized 
with arches of finite dimensions by making the arch-ring 
contain the imaginary linear arch, and the joints ~\ to the 
curve of the same. Questions of friction and the resist- 
ance of the material of the voussoirs are reserved for a, 
succeeding chapter, (§ 344) in which will be advanced a 
more practical theory dealing with approximate linear 
arches or " equilibrium polygons " as they will then be 
called. Still, a study of exact linear arches is valuable on 
many accounts. By inverting the linear arches so far pre- 
sented we have the forms assumed by flexible and inexten- 
sible cords loaded in the same way, 



GRAPHICAL STATICS. 397 



CHAPTER VIII. 



ELEMENTS OF GRAPHICAL STATICS. 



324. Definition. — In many respects graphical processes 
have advantages over the purely analytical, which recom- 
mend their use in many problems where celerity is desired 
without refined accuracy. One of these advantages is that 
gross errors are more easily detected, and another that 
the relations of the forces, distances, etc., are made so 
apj>arent to the eye, in the drawing, that the general effect 
of a given change in the data can readily be predicted at 
a glance. 

Graphical Statics is the system of geometrical construc- 
tions by which problems in Statics may be solved by 
the use of drafting instruments, forces as well as distances 
being represented in amount and direction by lines on the 
paper, of proper length and position, according to arbi- 
trary scales ; so many feet of distance to the linear inch of 
paper, for example, for distances ; and so many pounds or 
tons to the linear inch of paper for forces. 

Of course results should be interpreted by the same 
scale as that used for the data. The parallelogram of 
forces is the basis of all constructions for combining and 
resolving forces. 

325. Force Polygons and Concurrent Forces in a Plane. — If a 
material point is in equilibrium under three forces P x P 2 
P 3 (in the same plane of course) Fig. 344, any one of them, 



398 



MECHANICS OF ENGINEERING. 




as Pj, must be equal and opposite to R the resultant of 
the other two (diagonal of their parallelogram). If now 
we lay off to som convenient scale a line in Fig. 345 = 
P x and || to Px in Fig. 344 ; and then from the pointed end 

of P l a line equal and || to P 2 and 
laid off pointing the same ivay, we 
note that the line remaining to 
P close the triangle in Fig. 345 must 
be = and || to P 3 , since that tri- 
angle is nothing more than the 
left-hand half-parallelogram of 
fig. 345. Fig. 344. Also, in 345, to close 
the triangle properly the directions of the arrows must 
be continuous Point to Butty round the periphery. Fig. 
345 is called a force polygor ; of three sides only in this 
case. By means of it, given any two of the three forces 
which hold the point in equilibrium, the third can be 
found, being equal and || to the side necessary to " close '* 
the force polygon. 

Similarly, if a number of forces in a plane hold a mate- 
rial point in equilibrium, Fig. 346, their force polygon, 




Fig 344 




Fig. 346. 




Fig. 347, must close, whatever be the order in which its 
sides are drawn. For, if we combine P Y and P 2 into a re- 
sultant 0a t Fig. 346, then this resultant with P 3 to form a 
resultant Ob, and so on ; we find the resultant of P l9 P 2 ,P 3 , 
and P 4 to be Oc, and if a fifth force is to produce equilib- 
rium it must be equal and opposite to Oc, and would close 
the polygon OdabcO, in which the sides are equal and par- 



GRAPHICAL STATICS. 



399 



allel respectively to the forces mentioned. To utilize this 
fact we can dispense with all parts of the parallelograms in 
Fig. 346 except the sides mentioned, and then proceed as 
follows in Fig. 347 : 

If P 5 is the unknown force which is to balance the other 
four (i.e, is their anti-resultant), we draw the sides of the 
force polygon from A round to B, making each line paral- 
lel and equal to the proper force and pointing the same 
way ; then the line BA represents the required P 5 in 
amount and direction, since the arrow BA must follow 
the continuity of the others (point to butt). 

If the arrow BA were pointed at the extremity B, then 
it gives, obviously, the amount and direction of the result- 
ant of the four forces P x . . . P 4 . The foregoing shows 
that if a system of Concurrent Forces in a Plane is in equi- 
librium, its force polygon must close. 



326. Non-Concurrent Forces in a Plane. — Given a system of 
non-concurrent forces m a plane, acting on a rigid body, 
required graphic means of finding their resultant and anti- 
resultant ; also of expressing conditions of equilibrium. 
The resultant must be found in amount and direction ; and 
also in position (i.e., its line of action must be determined). 
E. g., Fig. 348 shows a curved rigid beam fixed in a vise 
at T, and also under the action of forces P x P 2 P 3 and P 4 
(besides the action of the vise); required the resultant of 

P^P^Ps, and P 4 . 
By the ordinary 
parallelogram of 
forces we com- 
bine P 1 and P 2 at 
a, the intersection 
of their lines of 
action, into a re- 
sultant P a ; then P a with P 3 at b, to form P b ; and finally P b 
with P 4 at c to form R c which is .*. the resultant required, 
i.e., of Pj . . . . P 4 ; and c . . . F is its line of action. 




400 



MECHANICS OF ENGINEERING. 




Fig. 349. 



The separate force triangles (half-parallelograms) by 
which the successive partial resultants P a , etc., were found, 
are again drawn in Fig. 349. Now since R c , acting in the 

line c.F, Fig. 348, 

is the resultant of 

/ Pj . . P 4 , it is plain 

, x x r* that a force B c ' 

equal to R c and act- 
ing along c . . P,but 
in the opposite di- 
rection, would balance the system P x . . . P 4 , (is their anti- 
resultant). That is, the forces P [ P 2 P 3 P 4 and RJ would 
form a system in equilibrium. The force RJ then, repre- 
sents the action of the vise T upon the beam. Hence re- 
place the vise by the force RJ acting in the line . . . P. . . c - 
to do which requires us to imagine a rigid prolongation of 
that end of the beam, to intersect P. . . c. This is shown in 
Fig. 350 where the whole beam is free, in equilibrium, under 
the forces shown, and in precisely the same state of stress, 
part for part, as in Fig. 348. Also, by combining in one 
force diagram, in Fig. 351, all the force triangles of Fig. 349 
(by making their common sides coincide, and putting RJ 
instead of R c , and dotting all forces other than those of 
Fig. 350), we have a figure to be interpreted in connection 
with Fig. 350. 



*-Y 




SPACE DIAGRAM 
FIG. 350. 



FORCE DIAGRAM 
Fig. 351. 



Here we note, first, that in the figure called a force-dia- 
gram, P Y P 2 P 3 P 4 and R c ' form a closed polygon and that 



GRAPHICAL STATICS. 401 

their arrows follow a continuous order, point to butt, 
around the perimeter ; which proves that one condition of 
equilibrium of a system of non-concurrent forces in a plane 
is that its force polygon must close. Secondly, note that ab 
is [| to Oa', and bc to Ob' ; hence if the force-diagram has 
been drawn (including the rays, dotted) in order to deter- 
mine the amount and direction of P c ', or any other one force, 
we may then find its line of action in the space-diagram, as 
follows: (N. B. — By space diagram is meant the figure show- 
ing to a true scale the form of the rigid body and the lines 
of action of the forces concerned). Through a, the intersec- 
tion of P Y and P 2 ,'draw a line || to Oa' to cut P 3 in some point 
b ; then through b a line || to Ob' to cut P 4 at some point c; cF 
drawn || to Oc' is the required line of action of P/, the anti- 
resultant of P lf P 2 , P 3 , and P 4 . 

abc is called an equilibrium polygon; this one having but 
two segments, ab and bc (sometimes the lines of action of P x 
and RJ may conveniently be considered as segments.) The 
segments of the equilibrium polygon are parallel to the respect- 
ive rays of the force diagram. 

Hence for the equilibrium of a system of non-concurrent 
forces in a plane not only must its force polygon close, 
but also the first and last segments of the corre- 
sponding equilibrium polygon must coincide with 
the resultants of the first two forces, and of the last 
two forces, respectively, of the system. E.g., ab coin- 
cides with the line of action of the resultant of P x and P 2 ; 
bc with that of P 4 and B' c . Evidently the equil. polygon 
will be different with each different order of forces in 
the force polygon or different choice of a pole, 0. But if 
the order of forces be taken as above, as they occur along . 
the beam, or structure, and the pole taken at the " butt " of 
the first force in the force polygon, there will be only one ; 
(and this one will be called the special equilibrium polygon 
in the chapter on arch-ribs, and the " true linear arch " in 
dealing with the stone arch.) After the rays (dotted in 
Fig. 351) have been added, by joining the pole to each 



402 



MECHANICS OF ENGINEERING. 



vertex with which it is not already connected, the final 
figure may be called the force diagram, ^ 

It may sometimes be convenient to give the name of 
rays to the two forces of the force polygon which meet 
at the pole, in which case the first and last segments of 
the corresponding equil. polygon will coincide with the 
lines of action of those forces in the space-diagram (as we 
may call the representation of the body or structure on 
which the forces act). This " space diagram " shows the 
real field of action of the forces, while the force diagram,, 
which may be placed in any convenient position on the 
paper, shows "the magnitudes and directions of the forces 
acting in the former diagram, its lines being interpreted 
on a scale of so many lbs. or tons to the inch of paper ; in 
the space-diagram we deal with a scale of so many feet to 
the inch of paper. 

We have found, then, that if any vertex or corner of the 
closed force polygon be taken as a pole, and rays drawn 
from it to all the other corners of the polygon, and a cor- 
responding equil. polygon drawn in the space diagram, the. 
first and last segments of the latter polygon must co-ineide 
with the first and last forces according to the order 
adopted (or with the resultants of the first two and last 
two, if more convenient to classify them thus). It remains, 
to utilize this principle. 



327. To Pind the Resultant of Several Forces in a Plane. — This 
might be done as in § 326, but since frequently a given set 
of forces are parallel, or nearly so, a special method will 
now be given, of great convenience in such cases. Fig. 352. 

Let P l P 2 and 
P 3 be the given 
forces whose 
resultant is re- 
quired. Let us 
first find their 
an^i - resultant,, 
or force which 
fig. 352. Fig. 353. will balance 




GRAPHICAL STATICS. 40 3 

them. This anti-resultant may be conceived as decom- 
posed into two components P and P' one of which, say P, 
is arbitrary in amount and position. Assuming P, then, 
at convenience, in the space diagram, it is required to find 
P'. The five forces must form a balanced system ; hence 
if beginning at O l9 Fig. 353, we lay off a line O x A = P by 
scale, then A\ = and || to P { , and so on (point to butt), the 
line BO x necessary to close the force polygon is = P' re- 
quired. Now form the corresponding equil. polygon in 
the space diagram in the usual way, viz.: through a the 
intersection of P and P 1 draw ab || to the ray L . . . 1 
(which connects the pole O x with the point of the last force 
mentioned). From b, where ab intersects the line of P 2 , 
draw be, || to the ray L . . 2, till it intersects the line of P 3 . 
A line mc drawn through c and || to the P' of the force 
diagram is the line of action of P f . 

Now the resultant of P and P f is the anti-resultant of 
P ly P 2 and P 3 ; .'. d, the intersection of the lines of P and 
P l , is a point in the line of action of the anti-resultant re- 
quired, while its direction and magnitude are given by the 
line BA in the force diagram ; for BA forms a closed poly- 
gon both with P x P 2 P 3 , and with PP f . Hence a line 
through d || to BA, viz., de, is the line of action of the anti- 
resultant (and hence of the resultant) of P l9 P 2 , P 3 . 

Since, in this construction, P is arbitrary, we may first 
choose O l9 arbitrarily, in a convenient position, i.e., in such 
a position that by inspection the segments of the result- 
ing equil. polygon shall give fair intersections and not 
pass off the paper. If the given forces are parallel the 
device of introducing the oblique P and P' is quite neces- 
sary. 

328. — The result of this construction may be stated as 
follows, (regarding Oa and cm as segments of the equil. 
polygon as well as ab and be): If any tivo segments of an 
equil polygon be prolonged, their intersection is a point in 
the line of action of the resultant of those forces acting at 



404 



MECHANICS OF ENGINEERING. 



the vertices intervening between the given segments. Here, 
the resultant of P x P 2 P 3 acts through d. 

329. Vertical Reaction of Piers, etc. — Fig. 354. Given the 
vertical forces or loads P x P 2 and P 3 acting on a rigid body 
(beam, or truss) which is supported by two piers having 
smooth horizontal surfaces (so that the reactions must be 
vertical), required the reactions V and V n of the piers. 
For an instant suppose V and V Q known; they are in 




Fig. 354. 

equil. with P x P 2 and P 3 . The introduction of the equal 
and opposite forces P and P' in the same line will not dis- 
turb the equilibrium. Taking the seven forces in the 
order P V P r P 2 P 3 V n and P', a force polygon formed with 
them will close (see (b) in Fig. where the forces which 
really lie on the same line are slightly separated). With 
0, the butt of P, as a pole, draw the rays of the force dia- 
gram OA, OB, etc. The corresponding equil. polygon 
begins at. a, the intersection of P and V in (a) (the space 
diagram), and ends at n the intersection of P' and V n . 
Join an. Now since P and P' act in the same line, an 
must be that line and must be || to P and P' of the force 
diagram. Since the amount and direction of P and P' are 
arbitrary, the position of the pole is arbitrary, while 
Pi, P 2 , and P 3 are the only forces known in advance in the 
force diagram. 

Hence V and V n may be determined as follows : Lay off 
the given loads P 1? P 2 , etc., in the order of their occur- 
rence in the space diagram, to form a " load-line " AD 



GRAPHICAL STATICS. 



405 



(see (b.) Fig. 354) as a beginning for a force-diagram ; take 
any convenient pole 0, draw the rays OA, OB, 00 and 
01). Then beginning at any convenient point a in the 
vertical line containing the unknown V , draw ab || to OA, 
be || to OB, and so on, until the last segment (dn in this 
case) cuts the vertical containing the unknown V n in some 
point n. Join an (this is sometimes called a closing line) 
and draw a || to it through 0, in the force -diagram. This 
last line will cut the " load-line " in some point n' , and 
divide it in two parts n' A and Dn', which are respectively 
V and V u required. 

Corollary. — Evidently, for a given system of loads, in given 
vertical lines of action, and for two given piers, or abut- 
ments, having smooth horizontal surfaces, the location of the 
point ?i A on the load line is independent of the choice of a 
pole. 

Of course, in treating the stresses and deflection of the 
rigid body concerned, P and P' are left out of account, as 
being imaginary and serving only a temporary purpose. 

330. Application of Foregoing Principles to a Roof Truss — 
Eig. 355. Wi and W 2 are wind pressures, P Y and P 2 are 
loads, while the remaining external forces, viz., the re- 






\w. 


/Vo 


\ 


/ ^ 


^w 


/ „-- 






3 

n 


\ x . ~~~ 


^_ 






\ V 


^-4* 


\ 


v 


\ 


\ 


\ 


4 — ^ 



6 H, 



400 MECHANICS OF ENGINEERING. 

actions, or supporting forces, V , V a and H ai may be found 
by preceding §§. (We here suppose that the right abut- 
ment furnishes all the horizontal resistance ; none at the 
left). 

Lay off the forces (known) W lt W 2 , P lf and P 2 in the 
usual way, to form a portion of the closed force polygon. 
To close the polygon it is evident we need only draw a 
horizontal through 5 and limit it by a vertical through 1. 
This determines H n but it remains to determine n' the 
point of division between V Q and V n . Select a convenient 
pole lf and draw rays from it to 1, 2, etc. Assume a con- 
venient point a in the line of V in the space diagram, and 
through it draw a line || to 0{i to meet the line of W x in 
some point b ; then a line || to X 2 to meet the line of W 2 
in some point c ; then through c || to 0$ to meet the line 
of P x in some point d ; then through d || to : 4 to meet the 
line of P 2 in some point e, (e is identical with d, since P 1 
and P 2 are in the same line) ; then ef II to X 5 to meet H n 
in some point/; then fg || to Ofi to meet V n in some 
point g. 

abcdefg is an equilibrium polygon corresponding to the 
pole X . 

Now join ag, the " closing-line," and draw a || to it 
through Ox to determine n', the required point of division 
between V and V n on the vertical 1 6. Hence V Q and V n 
are now determined as well as H Q . 

[The use of the arbitrary pole X implies the temporary 
employment of a pair of opposite and equal forces in the line 
ag, the amount of either being = 0^']. 

Having now all the external forces acting on the truss, 
and assuming that it contains no " redundant parts," i.e., 
parts unnecessary for rigidity of the frame-work, we proceed 
to find the pulls and thrusts in the individual pieces, on 
the following plan. The truss being pin- connected, no 
piece extending beyond a joint, and all loads being con- 
sidered to act at joints, the action, pull or thrust, of each 
piece on the joint at either extremity will be in the direction 
of the piece, i.e., in a hnoiun direction, and the pin of each 



GRAPHICAL STATICS. 407 

joint is in equilibrium under a system of concurrent forces 
consisting of the loads (if any) at the joint and the pulls 
or thrusts exerted upon it by the pieces meeting there. 
Hence we may apply the principles of § 325 to each joint 
in turn. See Fig. 356. In constructing and interpreting 
the various force polygons, Mr. K. H Bow's convenient 
notation will be used ; this is as follows : In the space 
diagram a capital letter [ABC, etc.] is placed in each tri- 
angular cell of the truss, and also in each angular space in 
the outside outline of the truss between the external forces 
and the adjacent truss-pieces. In this way we can speak of 
the force W x as the force BC, of W 2 as the force CE, the 
stress in the piece a/3 as the force CD, and so on. That 
is, the stress in any one piece can be named from the 
letters in the spaces bordering its two sides. Corresponding 
to these capital letters in the spaces of the space-dia- 
gram, small letters will be used at the vertices of the closed 
force-polygons (one polygon for each joint) in such a way 
that the stress in the piece CD, for example, shall be the 
force cd of the force polygon belonging to any joint in 
which that piece terminates ; the stress in the piece FG 
by the force fg in the proper force polygon, and so on. 

In Fig. 356 the whole truss is shown free, in equili- 
brium under the external forces. To find the pulls or 
thrusts (i.e., tensions or compressions) in the pieces, con- 
sider that if all but two of the forces of a closed force 
polygon are known in magnitude and direction, while the 
directions, only, of those two are known, the tcJiole force 
polygon may be drawn, thus determining the amounts of 
those two forces by the lengths of the corresponding 
sides. 

We must .-. begin with a joint where no more than two 
pieces meet, as at a ; [call the joints a, /9, r, d, and the cor- 
corresponding force polygons a', /3' etc. Fig. 356.] Hence 
at a' (anywhere on the paper) make ah || and = (by scale) 
to the known force AB (i.e., V ) pointing it at the upper end, 
and from this end draw be = and || to the known force BC 
(i.e., W x ) pointing this at the lower end. 



408 



MECHANICS OF ENGINEERING. 




Fig. 356. 

To close the polygon draw through c a || to the piece 
CD, and through a a || to AD ; their intersection deter- 
mines d, and the polygon is closed. Since the arrows 
must be point to butt round the periphery, the force with 
which the piece CD acts on the pin of the joint a is a 
force of an amount = cd and in a direction from c toward 
d ; hence the piece CD is in compression ; whereas the 
action of the piece DA upon the pin at a is from d toward 
a (direction of arrow) and hence DA is in tension. Notice 
that in constructing the force polygon «' a right-handed 
(or clock-wise) rotation has been observed in considering 
in turn the spaces ABC and D, round the joint a. A 
similar order will be found convenient in each of the other 
joints. 

Knowing now the stress in the piece CD, (as well as in 
DA) all but two of the forces acting on the pin at the joint 
ft are known, and accordingly we begin a force polygon, ft\ 
for that joint by drawing dc,= and || to the dc of polygon 
«', but pointed in the opposite direction, since the action of 
CD on the joint ft is equal and opposite to its action on 
the joint a (this disregards the weight of the piece). 
Through c draw ce = and || to the force CE (i.e., W 2 ) and 



GRAPHICAL STATICS. 409 

pointing the same way ; then ef, = and || to the load EF 
(i.e. Py) and pointing downward. Through f draw a || to 
the piece FG and through d, a || to the piece GD, and the 
polygon is closed, thus determining the stresses in the 
pieces FG and GD, Noting the pointing of the arrows, 
we readily see that FG is in compression while GD is in 
tension. 

Next pass to the joint d, and construct the polygon d' t 
thus determining the stress gh in GH and that ad in AD ; 
this last force ad should check with its equal and oppo- 
site ad already determined in polygon a'. Another check 
consists in the proper closing of the polygon j\ all of 
whose sides are now known. 

[A compound stress -diagram may be formed by super- 
posing the polygons already found in such a way as to 
make equal sides co-incide ; but the character of each 
stress is not so readily perceived then as when they are 
kept separate]. 

In a similar manner we may find the stresses in any pin- 
connected frame-work (in one plane and having no redun- 
dant pieces) under given loads, provided all the support- 
ing forces or reactions can be found. In the case of a 

braced-arch (truss) as 
shown in Fig. 357, hinged 
to the abutments at both 
ends and not free to slide 
laterally upon them, the 
reactions at and B de- 
Fig- 357. pend, in amount and direc- 

tion, not only upon the equations of Statics, but on the 
form and elasticity of the arch-truss. Such cases will be 
treated later under arch -ribs, or curved beams. 

332. The Special Equil. Polygon. Its Relation to the Stresses 
in the Rigid Body. — Reproducing Figs. 350 and 351 in Figs. 
358 and 359, (where a rigid curved beam is in equilibrium 
under the forces P lt P 2 , P 3 , P 4 and R\) we call a . . b . . c 




410 



MECHANICS OF ENGINEERING. 



the special equil. polygon because it corresponds to a force 
diagram in which the same order of forces has been ob- 
served as that in which they occur along the beam (from 
left to right here). From the relations between the force 




SPACE DIAGRAM 
Fig ..358. 



FORCE DIAGRAM 
Jig. 359. 



diagram and equil. polygon, this special equil. polygon in 
the space diagram has the following properties in connec- 
tion with the corresponding rays (dotted lines) in the force 
diagram. 

The stresses in any cross-section of the portion O'A of 
the beam, are due to P x alone ; those of any cross-section 
on AB to P l and P 2 , i.e., to their resultant P a , whose mag- 
nitude is given by the line Oa' in the force diagram, while 
its line of action is ab the first segment of the equil. poly- 
gon. Similarly, the stresses in BG are due to P b P 2 and 
P 3 , i.e., to their resultant i? b acting along the segment be, 
its magnitude being = Ob' in the force diagram. E.g., if 
the section at m be exposed, considering O'ABm as a free 
body, we have (see Fig. 360) the elastic stresses (or inter- 

6--TV — "rP* 

^ \v .m 

Fig. 360. Fig. 361. 

nal forces) at m balancing the exterior or " applied forces " 
P l5 P 2 and P 3 . Obviously, then, the stresses at m are just 




GKAPHICAL STATICS. 411 

the same as if i? b the resultant of P ]} P 2 and P 3 , acted upon 
an imaginary rigid prolongation of the beam intersecting 
be (see Fig. 361). B h might be called the " anti-stress-result- 
ant " for the portion BC of the beam. "We may.-, state 
the following : If a rigid body is in equilibrium under a sys- 
tem of Non-Concurrent Forces in a plane, and the special equi- 
librium polygon has been draiun, then each ray of the force 
diagram is the anti-stress-resultant of that portion of the beam 
which corresponds to the segment of the equilibrium polygon 
to ivhich the ray is parallel ; and its line of action is the seg- 
ment just mentioned. 

Evidently if the body is not one rigid piece, but com- 
posed of a ring of uncemented blocks (or voussoirs), it may 
be considered rigid only so long as no slipping takes place 
or disarrangement of the blocks ; and this requires that the 
" anti-stress-resultant " for a given joint between two 
blocks shall not lie outside the bearing surface of the 
joint, nor make too small an angle with it, lest tipping or 
slipping occur. For an example of this see Fig. 362, show- 
ing a line of three blocks in equilibrium under five forces. 

The pressure borne at the 
joint MN, is = i? a in the 
force -diagram and acts in 
the line ab. The con- 
struction supposes all 
the forces given except 
fig. 36-3. one, in amount and posi- 

tion, and that this one could easily be found in amount, as 
being the side remaining to close the force polygon, while 
its position would depend on the equil. polygon. But in 
practice the two forces P 1 and R' c are generally unknown, 
hence the point 0, or pole of the force diagram, can not 
be fixed, nor the special equil. polygon located, until other 
considerations, outside of those so far presented, are 
brought into play. In the progress of such a problem, as 
will be seen, it will be necessary to use arbitrary trial po- 
sitions for the pole 0, and corresponding trial equilibrium 
polygons. 




412 



MECHANICS OE EXGIXEEKQfG. 



CHAPTEE IX. 



GRAPHICAL STATICS OF VERTICAL FORCES. 



333. Remarks. — (With the exception of § 378 a) in prob- 
lems to be treated subsequently (either the stiff arch-rib, 
or the block-work of an arch-ring, of masonry) when the 
body is considered free all the forces holding it in equil. 
will be vertical (loads, due to gravity) except the reactions 
at the two extremities, as in Fig. 363 ; but for convenience 
each reaction will be replaced by its horizontal and verti- 
cal components (see Fig. 364). The two ZTs are of course 
equal, since they are the only horizontal forces in the 
system. Henceforth, aU equil. polygons under discussion iciU 
be understood to imply this kind of system of forces. F lt F 2 , 



us 



p f 2 f 3 1, 



/ 




Fig. 363. 



Fig. 364. 



A H 




rt 



FIG. 364a. 



etc., will represent the " loads " ; V and V Q the vertical 
components of the abutment reactions ; H the value of 
either horizontal component of the same. (We here sup- 
pose the pressures T and T n resolved along the horizon- 
tal and vertical.) 



GRAPHICAL STATICS. 



413 



334. Concrete Conception of an Equilibrium Polygon. — Any 
equilibrium polygon has this property, due to its mode 
of construction, viz.: If the ab and be of Fig. 358 were im- 
ponderable straight rods, jointed at b without friction, they 
would be in equilibrium under the system of forces there 
given. (See Fig. 364a). The rod ab suffers a compression 
equal to the i? a of the force diagram, Fig. 359, and be a 
compression = i? b . In some cases these rods might be in 
tension, and would then form a set of links playing the 
part of a suspension-bridge cable. (See § 44). 

335. Example of Equilibrium Polygon Drawn to Vertical Loads 
— Fig. 365. [The structure bearing the given loads is not 
shown, but simply the imaginary rods, or segments of an 
equilibrium polygon, which would support the given loads 
in equilibrium if the abutment points A and B, to which 
the terminal rods are hinged, were firm. In the present 
case this equilibrium is unstable since the rods form a 
standing structure ; but if they were hanging, the equilibri- 
um would be stable. Still, in the present case, a very light 
bracing, or a little friction at all joints would make the 
equilibrium stable. 




2 FT. TO INCH 



*»«.BS.TO INCH* vY" 



Fig. 365. 



Given three loads P lt P 2 , and P 3 , and two " abutment 
verticals " A' and B' ', in which we desire the equil. poly- 
gon to terminate, lay off as a " load-line," to scale, P lf P 2 , 
and P 3 end to end in their order. Then selecting any pole, 



414 MECHANICS OF ENGINEERING. 

0, draw the rays 01, 02, etc., of a force diagram (the V f & 
and P's, though really on the same vertical, are separated 
slightly for distinctness ; also the ZT's, which both pass 
through and divide the load-line into V and V n ). We 
determine a corresponding equilibrium polygon by draw- 
ing through A (any point in A') a line || to . . 1, to inter- 
sect Pi in some point b ; through b a || to . . 2, and so on> 
until B' the other abutment-vertical is struck in some 
point B. AB is the " abutment -line " or " closing -line." 

By choosing another point for 0, another equilibrium 
polygon would result. As to which of the infinite 
number (which could thus be drawn, for the given loads 
and the A' and B' verticals) is the special equilibrium poly- 
gon for the arch-rib or stone-arch, or other structure, on 
which the loads rest, is to be considered hereafter. In 
any of the above equilibrium polygons the imaginary 
series of jointed rods would be in equilibrium. 

336. Useful Property of an Equilibrium Polygon for Vertical. 
Loads. — (Particular case of § 328). See Fig. 366. In any 
equil. polygon, supporting vertical loads, consider as free 
any number of consecutive segments, or rods, with the 
loads at their joints, e. g., the 5th and 6th and portions of 

the 4th and 7th which, we sup- 
pose cut and the compressive 
forces in them put in, T\ and 
\^ T lt in order to consider 4 5 6 7 
as a free body. For equil., 
according to Statics, the lines 
D ip 6 \ of action of T 4 and T 7 (the com- 

fig. 366. pression in those rods) must in- 

tersect in a point, 0, in the line of action of the resultant 
of P 4 , P 5 , and P 6 ; i.e., of the loads occurring at the inter- 
vening vertices. That is, the point C must lie in the ver- 
tical containing the centre of gravity of those loads. Since 
the position of this vertical must be independent of the- 
particular equilibrium polygon used, any other (dotted 
lines in Fig. 366) for the same loads will give the same re- 



/ 1 


-v £iv^- 






'A^"'l 




/ 1 ' 

.- 1 — 

1 




f P* J 


1 P B 



GRAPHICAL STATICS. 415 

suits. Hence the vertical CD, containing the centre of 
gravity of any number of consecutive loads, is easily found 
by drawing the equilibrium polygon corresponding to 
any convenient force diagram having the proper load-line. 
This principle can be advantageously applied to finding 
a gravity -line of any plane figure, by dividing the latter 
into parallel strips, whose areas may be treated as loads 
applied in their respective centres of gravity. If the strips 
are quite numerous, the centre of gravity of each may be 
considered to be at the centre of the line joining the mid- 
dles of the two long sides, while their areas may be taken 
as proportional .to the lengths of the lines drawn through 
these centres of gravity parallel to the long sides and lim- 
ited by the end-curves of the strips. Hence the " load- 
line " of the force diagram may consist of these lines, or 
of their halves, or quarters, etc., if more convenient. 



USEFUL RELATIONS BETWEEN FORCE DIA- 
GRAMS AND EQUILIBRIUM POLYGONS. 

(for vertical loads.) 

337. Resume of Construction. — Fig. 367. Given the loads 
P l3 etc., their verticals, and the two abutment verticals A' 
and B', in which the abutments are to lie ; we lay off a 
load-line 1 ... 4, take any convenient pole, 0, for a force- 
diagram and complete the latter. For a corresponding 
equilibrium polygon, assume any point A in the vertical 
A\ for an abutment, and draw the successive segments 
Al, 2, etc., respectively parallel to the inclined lines of the 
force diagram (rays), thus determining finally the abut- 
ment B, in B\ which (B) will not in general lie in the hor- 
izontal through A. 

Now join AB, calling AB the abutment-line, and draw a 
parallel to it through 0, thus fixing the point n on the 



416 



MECHANICS OF ENGINEERING. 




Fig. 36 



load-line. This point n', as above determined, is indepen- 
dent of the location of the pole, 0, (proved in § 329) and 
divides the load-line into two portions ( V = 1 . . . n', and 
V' n = n f . . . 4) which are the vertical pressures which two 
supports in the verticals A' and B' would sustain if the 
given loads rested on a horizontal rigid bar, as in Fig. 368. 

See § 329. Hence to find the point n' we may use any 
convenient pole 0. 

[N. B.— The forces V and V Q of Fig. 367 are not identi- 
cal with V' and V' n , but may be obtained by dropping a 
"| from to the load-line, thus dividing the load-line 
into two portions which are V (upper portion) and V n . 
However, if A and B be connected by a tie-rod, in Fig. 
367, the abutments in that figure will bear vertical press- 
ures only and they will be the same as in Fig. 368, while 
the tension in the tie-rod will be = On'.] 

338. Theorem. — The vertical dimensions of any two equili- 
brium polygons, drawn to the same loads, load-verticals, and 
abutment-verticals, are inversely proportional to their H's (or 
"pole distances "). We here regard an equil. polygon and 
its abutment-line as a closed figure. Thus, in Fig. 369, 
we have two force-diagrams (with a common load-line, for 
convenience) and their corresponding equil. polygons, for 
the same loads and verticals. From § 337 we know that 
On' is || to AB and n' is || to A B . Let CD be any ver- 
tical cutting the first segments of the two equil. polygons. 



GRAPHICAL STATICS. 



417 



Denote the intercepts thus determined by z' and z' ot respect- 
ively. From the 
parallelisms just 
mentioned, and 
others more famil- 
^ iar, we have the 
triangle In' sim- 
ilar to the triangle 
Az' (shaded), and 
the triangle o ln' 
similar to the tri- 
angle A z£, Hence 



c 


P, 


1 


2- 


z 


p * y- 


r 








the proportions between j In' z 



"bases and altitudes 



=— and 



In' 



h ) 



{ H h H 

.'. z' : z' : : H : H. The same kind of proof may easily 

be applied to the vertical intercepts in any other segments, 
e. g., z" and z" . Q. E. D. 

339. Corollaries to the foregoing. It is evident that : 
(1.) If the pole of the force-diagram be moved along a 
vertical line, the equilibrium polygon changing its form 
in a corresponding manner, the vertical dimensions of the 
equilibrium polygon remain unchanged ; and 

(2.) If the pole move along a straight line which con- 
tains the point n' t the direction of the abutment-line 
remains constantly parallel to the former line, while the 
vertical dimensions of the equilibrium polygon change in 
inverse proportion to the pole distance, or H, of the force- 
diagram. [R is the 1 distance of the pole from the load- 
line, and is called the pole-distance]. 

§ 340. Linear Arch as Equilibrium Polygon. — (See § 316.) 
If the given loads are infinitely small with infinitely small 
horizontal spaces between them, any equilibrium polygon 
becomes a linear arch. Graphically we can not deal with 
these infinitely small loads and spaces, but from § 336 it 
is evident that if we replace them, in successive groups, 



418 



MECHANICS OF ENGINEERING. 



by finite forces, each of which 

IUUIU .HH.UH 




Fig. 370. 



= the sum of those com- 
posing one group and is 
applied through the cen- 
tre of gravity of that 
group, we can draw an 
equilibrium polygon 
whose segments will be 
tangent to the curve of 
the corresponding linear 
arch, and indicate its posi- 
tion with sufficient exactness for practical purposes. (See 
Fig. 370). The successive points of tangency A, m, n, etc., 
lie vertically under the points of division between the 
groups. This relation forms the basis of the graphical 
treatment of voussoir, or blockwork, arches. 

341. To Pass an Equilibrium Polygon Through Three Arbitrary 
Points. — (In the present case the forces are vertical. For 
a construction dealing with any plane system of forces see 
construction in § 378$.) Given a system of loads, it is re- 
quired to draw 
\ ~ y ~* an equilibrium 

them through 
any three points, 
two of which 
may be consid- 
ered as abut- 
ments, outside of the load-verticals, the third point being 
between the verticals of the first two. See Fig. 371. The 
loads P l9 etc., are given, with their verticals, while A, p, 
and B are the three points. Lay off the load-line, and 
with any convenient pole, O x , construct a force-diagram, 
then a corresponding preliminary equilibrium polygon 
beginning at A. Its right abutment B u in the vertical 
through B, is thus found. O x n' can now be drawn || to AB V 
to determine n'. Draw n'O II to BA. The pole of the 
required equilibrium polygon must lie on n'O (§ 337)* 




Fig. 371. 



GEAPHICAL STATICS. 



419 



Draw a vertical through p. The H of the required equili- 
brium polygon must satisfy the proportion H : H x : : rs : 
pm. (See § 338). Hence construct or compute H from 
the proportion and draw a vertical at distance H from 
the load-line (on the left of the load-line here) ; its inter- 
section with n' gives the desired pole, for which a 
force diagram may now be drawn. The corresponding 
equilibrium polygon beginning at the first point A will 
also pass through p and B ; it is not drawn in the figure. 




342. Symmetrical Case of the Foregoing Problem.— If two 
points A and B are on a level, the third, p, on the middle 
vertical between them ; and the loads (an even number) 
symmetrically disposed both in position and magnitude, about 
p, we may proceed more simply, as follows : (Fig. 372). 

From symmetry n' 
must occur in the mid- 
dle of the load-line, of 
which we need lay off 
only the upper half. 
Take a convenient pole 
Ox, in the horizontal 
through n', and draw a half force diagram and a corres- 
ponding half equilibrium polygon (both dotted). The up- 
per segment be of the latter must be horizontal and being 
prolonged, cuts the prolongation of the first segment in a 
point d, which determines the vertical CD containing the 
centre of gravity of the loads occurring over the half -span 
on the left. (See § 336). In the required equilibrium poly- 
gon the segment containing the point p must be horizon- 
tal, and its intersection with the first segment must lie in 
CD. Hence determine this intersection, C, by drawing the 
vertical CD and a horizontal through p ; then join CA, 
which is the first segment of the required equil. polygon. 
A parallel to CA through 1 is the first ray of the corres- 
ponding force diagram, and determines the pole on the 
horizontal through n'. Completing the force diagram for 



420 



MECHANICS OF ENGINEERING. 



this pole (half of it only here), the required equil. poly- 
gon is easily finished afterwards. 

343. To Find a System of Loads Under Which a Given Equi- 
librium Polygon Would be in Equilibrium, — Fig. 373. Let AB 
be the given equilibrium polygon. Through any point 
as a pole draw a parallel to each 
segment of the equilibrium polygon. 
Any vertical, as V, cutting these 
lines will have, intercepted upon it, 
N> ^3^ a load-line 1, 2, 3, whose parts 1 . . 2, 
v l 2 . . 3, etc., are proportional to the 
fig. 373. successive loads which, placed on 

the corresponding joints of the equilibrium polygon would 
be supported by it in equilibrium (unstable). 

One load may be assumed and the others constructed. 
A hanging, as well as a standing, equilibrium polygon 
may be dealt with m like manner, but will be in stable equi- 
librium. The problem in § 44 may be solved in this way. 




AKCHES OF MASONRY. 



421 



CHAPTEE X. 



RIGHT ARCHES OF MASONRY. 



344.— In an ordinary "right" stone-arch (i.e., one in 
which the faces are "1 to the axis of the cylindrical soffit, 
or under surface), the successive blocks forming the arch- 
ring are called voussoirs, the joints between them being 
planes which, prolonged, meet generally in one or more 
horizontal lines; e.g., those of a three-centred arch in three 
II horizontal lines ; those of a circular arch in one, the axis 
of the cylinder, etc. Elliptic arches are sometimes used. The 
inner concave surface is called the soffit, to which the radiat- 
ing joints between the voussoirs are made perpendicular. 
The curved line in which the soffit is intersected by a plane 





/ 




ClIbWN • 






1/ 


70- 


RISE 

! 








A 




1 
i 

i 


SPAN g 













Fig. 374. 



T to the axis of the arch is the Intrados. The curve in the 
same plane as the intrados, and bounding the outer ex- 
tremities of the joints between the voussoirs, is called the 
Extrados. 

Fig. 374 gives other terms in use in connection with a 



422 



MECHANICS OF ENGINEERING. 



stone arch, and explains those already given. 
" springing-line." 



AB is the 



345, Mortar and Friction. — As common mortar hardens 
very slowly, no reliance should be placed on its tenacity 
as an element of stability in arches of any considerable 
size ; though hydraulic mortar and thin joints of ordinary 
mortar can sometimes be depended on. Friction, however, 
between the surfaces of contiguous voussoirs, plays an 
essential part in the stability of an arch, and will there- 
fore be considered. 

The stability of voussoir-arches must .V be made to 
depend on the resistance of the voussoirs to compresssion 
and to sliding upon each other ; as also of the blocks 
composing the piers, the foundations of the latter being 
firm. 

346. Point of Application of the Resultant Pressure between 
two consecutive voussoirs ; (or pier blocks). Applying 
Navier's principle (as in flexure of beams) that the press- 
ure per unit area on a joint varies uniformly from the 
extremity under greatest compression to the point of least 
compression (or of no compression); and remembering 
that negative pressures (i.e., tension) can not exist, as they 
might in a curved beam, we may represent the pressure 
per unit area at successive points of a joint (from the intra- 
dos toward the extrados, or vice versa) by the ordinates of 
a straight line, forming the surface of a trapezoid or tri- 
angle, in which figure the foot of the ordinate of the cen- 
tre of gravity is the point of application of the resultant 
^pressure. Thus, where the least compression is supposed 




*,^0f 





Fig. 



Fir,. 37G. 



Fig. 377. 




Fig. 37 



MASONRY ARCHES. 



423 



-to occur at the intrados A, Fig. 375, the pressures vary as 
ihe ordinates of a trapezoid, increasing to a maximum value 
at B y in the extrados. In Fig. 376, where the pressure is zero 
at B, and varies as the ordinates of a triangle, the result- 
ant pressure acts through a point one-third the joint- 
length from A. Similarly in Fig. 377, it acts one-third 
the joint-length from B. Hence, when the pressure is not 
zero at either edge the resultant pressure acts within the 
middle third of the joint. Whereas, if the resultant press- 
ure falls without the middle third, it shows that a portion 
Am of the joint, see Fig. 378, receives no pressure, i.e., the 
joint tends to open along Am. 

Therefore that no joint tend to open, the resultant press- 
ure must fall tvithin the middle third. 

It must be understood that the joint surfaces here dealt 
with are rectangles, seen edgewise in the figures. 

347. Friction. — By experiment it has been found the 
angle of friction (see § 156) for two contiguous voussoirs 
of stone or brick is about 30° ; i.e., the coefficient of fric- 
tion is f = tan. 30°. Hence if the direction of the press- 
ure exerted upon a voussoir by its neighbor makes an 
angle a less than 30° with the normal to the joint surface, 
there is no danger of rupture of the arch by the sliding 
of one on the other. (See Fig. 379). 

348. Resistance to Crushing. — When the resultant pressure 
falls at its extreme allowable limit, viz. : the edge of the 
middle third, the pressure per 
.unit of area at n, Fig. 380, is 
double the mean pressure per 
unit of area. Hence, in de- 
signing an arch of masonry, 
we must be assured that at 
every joint (taking 10 as a 
factor of safety) 

( Double the mean press- ) mugt be legg than y Q 
\ ure per unit oi area \ ' u 




Fig. 37 



424 MECHANICS OF ENGINEERING. 

G being the ultimate resistance to crushing, of the material 
employed (§ 201) (Modulus of Crushing). 

Since a lamina one foot thick will always be considered 
in what follows, careful attention must be paid to the units 
employed in applying the above tests. 

Example. — If a joint is 3 ft. by 1 foot, and the resultant 
pressure is 22.5 tons the mean pressure per sq. foot is 

p=22.5^3=7.5 tons per sq. foot 

.*. its double=15 tons per sq. foot=208.3 lbs. sq. inch, 
which is much less than 1 / 10 of G for most building stones ;. 
see § 203, and below. 

At joints where the resultant pressure falls at the middle, 
the max. pressure per square inch would be equal to the 
mean pressure per square inch ; but for safety it is best to 
assume that, at times, (from moving loads, or vibrations) 
it may move to the edge of the middle third, causing the 
max. pressure to be double the mean (per square inch). 

Gen. Gillmore's experiments in 1876 gave the following 
results, among many others : 

NAME OF BUILDING STONE. C IN LBS. PER SQ. INCH. 

Berea sand-stone, 2-inch cube, - 8955 

4 " " - 11720 

Limestone, Sebastopol, 2-inch cube - 1075 

Marble, Yermont, 2-inch cube, - > 8000 to 13000 

Granite, New Hampshire, 2-inch cube, 15700 to 24000 

349. The Three Conditions of Safe Equilibrium for an arch of 
uncemented voussoirs. 

Recapitulating the results of the foregoing paragraphs, 
we may state, as follows, the three conditions which must 
be satisfied at every joint of arch -ring and pier, for each 
of any possible combination of loads upon the structure : 

(1). The resultant pressure must pass within the middle- 
third. 

(2). The resultant pressure must not make an angle > 
30° with the normal to the joint. 

(3). The mean pressure per unit of area on the surface* 



ARCH OF MASONRY. 



425 




of the joint must not exceed 1 / 20 of the Modulus of crush- 
ing of the material. 

350. The True Linear- Arch, or Special Equilibrium Polygon ; 

and the resultant pressure at any joint. Let the weight 
of each voussoir and its load be represented by a vertical 
force passing through the centre of gravity of the two, as 
in Fig. 381. Taking any 
two points A and B, A 
being in the first joint and 
B in the last ; also a third 
point, p, in the crown 
joint (supposing such to 
be there, although gener- 
ally a key-stone occupies 
the crown), through these pig. 38i. 

three points can be drawn [§ 341] an equilibrium polygon 
for the loads given ; suppose this equil. polygon nowhere 
passes outside of the arch-ring (the arch-ring is the por- 
tion between the intrados, mn, and the (dotted) extrados 
m'n') intersecting the joints at b, c, etc. Evidently if such 
be the case, and small metal rods (not round) were insert- 
ed at A, b, c, etc., so as to separate the arch-stones slight- 
ly, the arch would stand, though in unstable equilibrium, 
the piers being firm ; and by a different choice of A, p, and 
By it might be possible to draw other equilibrium poly- 
gons with segments cutting the joints within the arch- 
ring, and if the metal rods were shifted to these new inter- 
sections the arch would again stand (in unstable equilib- 
rium). 

In other words, if an arch stands, it may be possible to 
draw a great number of linear arches within the limits of 
the arch-ring, since three points determine an equilibrium 
polygon (or linear arch) for given loads. The question 
arises then : ivhich linear arch is the locus of the actual re- 
sultant pressures at the successive joints ? 

[Considering the arch-ring as an elastic curved beam 
inserted in firm piers (i.e., the blocks at the springing-line 



426 MECHANICS OF ENGINEERING. 

are incapable of turning) and having secured a close fit at 
all joints before the centering is lowered, the most satisfac- 
tory answer to this question is given in Prof. Greene's 
" Arches," p. 131 ; viz., to consider the arch-ring as an 
arch rib of fixed ends and no hinges ; see § 380 of next 
chapter ; but the lengthy computations there employed 
(and the method demands a simple algebraic curve for the 
arch) may be most advantageously replaced by Prof. 
Eddy's graphic method (" New Constructions in Graphical 
Statics," published in Yan Nostrand's Magazine for 1877), 
which applies to arch curves of any form. 

This method will be given in a subsequent chapter, on 
Arch Eibs, or Curved Beams ; but for arches of masonry a 
much simpler procedure is sufficiently exact for practical 
purposes and will now be presented]. 

If two elastic blocks 
of an arch-ring touch at 
one edge, Fig. 382, their 
adjacent sides making a 
small angle with each 
Fig. 382. fig. 383. other, and are then grad- 

ually pressed more and more forcibly together at the edge 
m, as the arch-ring settles, the centering being gradually 
lowered, the surface of contact becomes larger and larger, 
from the compression which ensues (see Fig. 383); while 
the resultant pressure between the blocks, first applied at 
the extreme edge m, has now probably advanced nearer the 
middle of the joint in the mutual adjustment of the arch- 
stones. "With this in view we may reasonably deduce the 
following theory of the location of the true linear arch 
(sometimes called the " line of pressures " and " curve of 
pressure ") in an arch under given loading and with^rra 
piers. (Whether the piers are really unyielding, under the 
oblique thrusts at the springing-line, is a matter for sub- 
sequent investigation. 

351. Location of the True Linear Arch. — Granted that the 
voussoirs have been closely fitted to each other over the 





ARCH OF MASONRY. 427 

centering (sheets of lead are sometimes used in the joints 
to make a better distribution of pressure); and that the 
piers are firm ; and that the arch can stand at all without 
the centering ; then we assume that in the mutual accom- 
modation between the voussoirs, as the centering is low- 
ered, the resultant of the pressures distributed over any 
joint, if at first near the extreme edge of the joint, advances 
nearer to the middle as the arch settles to its final posi- 
tion of equilibrium under its load ; and hence the follow- 
ing 

352. Practical Conclusions. 

I. If for a given arch and loading, with firm piers, an 
equilibrium polygon can be drawn (by proper selection of 
the points A, p, and B, Fig. 381) entirely within the mid- 
dle third of the arch ring, not only will the arch stand, but 
the resultant pressure at every joint will be within the 
middle third (Condition 1, § 349) ; and among all possible 
equilibrium polygons which can be drawn within the mid- 
dle third, that is the " true " one which most nearly coin- 
cides with the middle line of the arch-ring. 

II. If (with firm piers, as before) no equilibrium poly- 
gon can be drawn within the middle third, and only one 
within the arch-ring at all, the arch may stand, but chip- 
ping and spawling are likely to occur at the edges of the 
joints. The design should .'. be altered. 

III. If no equilibrium polygon can be drawn within 
the arch-ring, the design of either the arch or the loading 
must be changed ; since, although the arch may stand, 
from the resistance of the spandrel walls, such a stability 
must be looked upon as precarious and not countenanced 
in any large important structure. (Yery frequently, in 
small arches of brick and stone, as they occur in buildings, 
the cement is so tenacious that the whole structure is vir- 
tually a single continuous mass). 

When the " true " linear arch has once been determined, 
"the amount of the resultant pressure on any joint is given 
by the length of the proper ray in the force diagram. 



428 



MECHANICS OF ENGINEERING. 



ARRANGEMENT OF DATA FOR GRAPHIC 
TREATMENT. 

353. Character of Load. — In most large stone arch bridges 
the load (permanent load) does not consist exclusively of 
masonry up to the road-way but partially of earth filling 
above the masonry, except at the faces of the arch where 
the spandrel walls serve as retaining walls to hold the 
earth. (Fig. 384). If the intrados is a half circle or half- 




Fig. 384. 



Fig. 385. 



ellipse, a compactly -built masonry backing is carried up 
beyond the springing-line to AB about 60° to 45° from the 
crown, Fig. 385 ; so that the portion of arch ring below 
AB may be considered as part of the abutment, and thus 
AB is the virtual springing-line, for graphic treatment. 

Sometimes, to save filling, small arches are built over 
the haunches of the main arch, with earth placed over 
them, as shown in Fig. 386. In any of the preceding cases 




Fig. 386. 



Fig. 387. 



it is customary to consider that, on account of the bond- 
ing of the stones in the arch shell, the loading at a given 
distance from the crown is uniformly distributed over the 
width of the roadway. 



ARCHES OF MASONRY. 429 

354. Reduced Load-Contour. — In the graphical discussion 
of a proposed arch we consider a lamina one foot thick, 
this lamina being vertical and "| to the axis of the arch ; 
i.e., the lamina is || to the spandrel walls. For graphical 
treatment, equal areas of the elevation (see Fig. 387) of 
this lamina must represent equal weights. Taking the 
material of the arch-ring as a standard, we must find for 
each point p of the extrados an imaginary height z of the 
arch-ring material, which would give the same pressure 
(per running horizontal foot) at that point as that due to 
the actual load above that point. A number of such or- 
dinates, each measured vertically upward from the extra- 
dos determine points in the "Reduced Load-Contour," i.e., 
the imaginary line, AM, the area between which and the 
extrados of the arch -ring represents a homogeneous load 
of the same density as the arch-ring, and equivalent to the 
actual load (above extrados), vertical by vertical, 

355. Example of Reduced Load-Contour. — Fig. 388. Given 
an arch-ring of granite (heaviness = 170 lbs. per cubic 
foot) with a dead load of rubble (heav. = 140) and earth 
(heav. = 100), distributed as in figure. At the point p, of 
the extrados, the depth 5 feet of rubble is equivalent to a 
depth of [Ji2. x5]=4.1 ft. of granite, while the 6 feet of earth 
is equivalent to [J^x6]=3.5 feet of granite. Hence the 
Reduced Load-Contour has an ordinate, above p, of 7.6 feet. 
That is, for each of several points of the arch -ring extrados 
reduce the rubble ordinate in the ratio of 170 : 140, and 
the earth ordinate in the ratio 170 : 100 and add the re- 
sults, setting off the sum vertically from the points in the 
extrados*. In this way Fig. 389 is obtained and the area 



♦This is most conveniently done by graphics, thus : On a right-line set off 17 equal 
parts (of any convenient magnitude.) Call this distance OA. Through draw another 
right line at any convenient angle (30° to 60°) with OA, and on it from O 

set off OB equal to 14 (for the rubble ; or 10 for the earth) of the same equal 
parts. Join AB. From toward A set off* all the rubble ordinates to be reduced, 
teach being set off from 0) and through the other extremity of each draw a line par- 
allel to AB. The reduced ordinates will be the respective lengths, from 0, along 0B 9 
*fco the intersections of these parallels with OB. 

* With the dividers. 



430 



MECHANICS OF ENGINEERING. 



HI 




Fig. 




there given is to be treated as representing homogeneous 
granite one foot thick. This, of course, now includes the 
arch-ring also. AB is the " reduced load-contour." 

356. Live Loads. — In discussing a railroad arch bridge 
the " live load " (a train of locomotives, e.g., to take an ex- 
treme case) can not be disregarded, and for each of its po- 
sitions we have a separate Reduced Load-Contour. 

Example. — Suppose the arch of Fig. 388 to be 12 feet 
wide (not including spandrel walls) and that a train of lo- 
comotives weighing 3,000 lbs. per running foot of the track 
covers one half of the span. Uniformly distributed later- 
ally over the width, 12 ft., this rate of loading is equiva- 
lent to a masonry load of one foot high and a heaviness of 
250 lbs. per cubic ft., i.e., is equivalent to a height of 1.4 
ft. of granite masonry [since |^ X 1.0=1.4] over the half 
span considered. Hence from Fig. 390 we obtain Fig. 391 
in an obvious manner. Fig. 391 is now ready for graphic 
treatment. 




Fig. 390. Fig. 391. 

357. Piers and Abutments. — In a series of equal arches 
the pier between two consecutive arches bears simply the 
weight of the two adjacent semi-arches, plus the load im- 



ARCHES OF MASONRY. 



431 



mediately above the pier, and .*. does not need to be as 
large as the abutment of the first and last arches, since 
these latter must be prepared to resist the oblique thrusts 
of their arches without help from the thrust of another on 
the other side. 

In a very long series of "arches it is sometimes customary 
to make a few of the intermediate piers large enough to 
act as abutments. These are called " abutment piers," and 
in case one arch should fall, no others would be lost except 
those occurring between the same two abutment piers as 
the first. See Fig. 392. A is an abutment-pier. 



no-poo 



Fiu. 8!tt. 



GRAPHICAL TREATMENT OF ARCH. 

358. — Having found the " reduced load-contour," as in 
preceding paragraphs, for a given arch and load, we are 
ready to proceed with the graphic treatment, i.e., the first 
given, or assumed, form and thickness of arch-ring is to be 
investigated with regard to stability. It may be necessary 
to treat, separately, a lamina under the spandrel wall, and 
one under the interior loading. The constructions are 
equally well adapted to arches of all shapes, to Gothic as 
well as circular and elliptical. 

359.— Case L Symmetrical Arch and Symmetrical Loading. — 
(The " steady " (permanent) or " dead " load on an arch is 
usually symmetrical). Fig. 393. From symmetry we need 

o 

T 




Fig. 393. 



Fig. 394. 



Fig. 395. 



432 MECHANICS OF ENGINEERING. 

deal with only one half (say the left) of the arch and load. 
Divide this semi-arch and load into six or ten divisions 
by vertical lines ; these divisions are considered as trape- 
zoids and should have the same horizontal width = o (a 
convenient whole number of feet) except the last one, LKN> 
next the abutment, and this is a pentagon of a different 
width &j, (the remnant of the horizontal distance LC), The 
weight of masonry in each division is equal to (the area 
of division) X (unity thickness of lamina) X (weight of a cu- 
bic unit of arch-ring). For example for a division having 
an area of 20 sq. feet, and composed of masonry weighing 
160 lbs. per cubic foot, we have 20x1x160=3,200 lbs., 
applied through the centre of gravity of the division. 
The area of a trapezoid, Fig. 394, is j4b{hi~\-h)> an d its cen- 
tre of gravity may be found, Fig. 395, by the construction 
of Prob. 6, in § 26 ; or by § 27a. The weight of the pen- 
tagon LN, Fig. 393, and its line of .application (through 
centre of gravity) may be found by combining results for 
the two trapezoids into which it is divided by a vertical 
through K. See § 21. 

Since the weights of the respective trapezoids {except, 
ing LN) are proportional to their middle vertical in- 
tercepts [such as j4(hi-\-7h) Fig. 394] these intercepts (trans- 
ferred with the dividers) may be used directly to form the 
load-line, Fig. 396, or proportional parts of them if more 
convenient. The force scale, which this implies, is easily 
computed, and a proper length calculated to represent the 
weight of the odd division LN ; i.e., 1 ... 2 on the load- 
line. 

Now consider A, the middle point of the abutment joint, 
Fig. 396, as the starting point of an equilibrium polygon 
(or abutment of a linear arch) for a given loading, and re- 
quire that this equilibrium polygon shall pass through p, 
the middle of the crown joint, and through the middle of 
the abutment joint on the right (not shown in figure). 

Proceed as in § 342, thus determining the polygon Ap 
for the half -arch. Draw joints in the arch -ring through 
those points where the extrados is intersected by the ver- 



ABCHES OF MASONRY. 



433 




Fig. 39G. 



Fig. 397 



tical separating the divisions (not the gravity verticals). 
The points in which these joints are cut by the segments 
of the equilibrium polygon, Fig. 397, are (very nearly, if 
the joint is not more than 60° from p, the crown) the points 
of application in these joints, respectively, of the resultant 
pressures on them, (if this is the " true linear arch " for 
this arch and load) while the amount and direction of each 
such pressure is given by the proper ray in the force -dia- 
gram. 

If at any joint so drawn the linear arch (or equilibrium 
polygon) passes outside the middle third of the arch-ring, 
the point A, or p, (or both) should be judiciously moved 
(within the middle third) to find if possible a linear arch 
which keeps within limits at all joints. If this is found 
impossible, the thickness of the arch -ring may be increased 
at the abutment (giving a smaller increase toward the 
crown) and the desired result obtained ; or a change in the 
distribution or amount of the loading, if allowable, may 
gain this object. If but one linear arch can be drawn 
within the middle third, it may be considered the " true " 
one ; if several, the one most nearly co-inciding with the 
middles of the joints (see §§ 351 and 352) is so considered. 



360.— Case II. TJnsymmetrical Loading on a Symmetrical Arch; 
(e.g., arch with live load covering one half -span as in Figs. 
390 and 391). Here we must evidently use a full force 
diagram, and the full elevation of the arch -ring and load. 



434 



MECHANICS OF ENGINEERING. 



See Fig. 398. Select three points A, p, and B, as follows, 
to determine a trial equilibrium polygon : 

Select A at the loiuer limit of the middle third of the 




Fig. 398. 

abutment-joint at the end of the span which is the more-, 
heavily -loaded ; in the other abutment-joint take B at the 
upper limit of the middle third ; and take p in the middle 
of the crown-joint. Then by § 341 draw an equilibrium 
polygon (i.e., a linear arch) through these three points for 
the given set of loads, and if it does not remain within the 
middle third, try other positions for A, p, and B, within 
the middle third. As to the " true linear arch " alterations 
of the design, etc., the same remarks apply as already 
given in Case I. Very frequently it is not necessary to 
draw more than one linear arch, for a given loading, for 
even if one could be drawn nearer the middle of the arch- 
ring than the first, that fact is most always apparent on 
mere inspection, and the one already drawn (if within 
middle third) will furnish values sufficiently accurate for 
the pressures on the respective joints, and their direction 
angles. 

360. — The design for the arch-ring and loading is not 
to be considered satisfactory until it is ascertained that for 
the dead load and any possible combination of live-load 
(in addition) the pressure at any joint is 



ARCHES OF MASONRY. 435 

(1.) Within the middle third of that joint ; 

(2.) At an angle of < 30° with the normal to joint- 
surface. 

(3.) Of a mean pressure per square inch not > than 1 / 20 
of the ultimate crushing resistance. (See § 348.) 

§361. Abutments. — The abutment should be compactly 
and solidly built, and is then treated "as a single rigid mass. 
The pressure of the lowest voussoir upon it (considering 
a lamina one foot thick) is given by the proper ray of the 
force diagram (0 .. l,e. g., in Fig. 396) in amount and direc- 
tion. The stability of the abutment will depend on the 
amount and direction of the resultant obtained by com- 
bining that pressure P a with the weight G of the abutment 
and its load, see Fig. 399. Assume a probable width RS 

m for the abutment and compute the weight G 

of the corresponding abutment OBRS and 

J MNB 0, and find the centre of gravity of the 

c . ""a whole mass C. Apply G in the vertical 

i / through (7, and combine it with P a at their in- 

y(b "^" n tersection D. The resultant P should not cut 

/l 1 the base RS in a point beyond the middle third 

P // (or, if this rule gives too massive a pier, take 

1 / / G such a width that the pressure per square 

!// inch at S shall not exceed a safe value as 

fig. 399. computed from § 362.) After one or two 
trials a satisfactory width can be obtained. 
We should also be assured that the angle PDG is less 
than 30°. The horizontal joints above RS should also be 
tested as if each were, in turn, the lowest base, and if 
necessary may be inclined (like mn) to prevent slipping. 
On no joint should the maximum pressure per square inch 
be > than 1 / 10 the crushing strength of the cement. Abut- 
ments of firm natural rock are of course to be preferred 
where they can be had. If water penetrates under an 
abutment its buoyant effort lessens the weight of the lat- 
ter to a considerable extent. 



436 



MECHANICS OF ENGINEEKING. 



362. Maximum Pressure Per Unit of Area "When the Resultant 
Pressure Falls at Any Given Distance from the Middle ; according 
to Navier's theory of the distribution of the pressure ; see 
§ 346. Case I. Let the resultant pressure P, Fig. 400, (a), 




Fig. 400. 



Fig. 401. 



fall within the middle third, a distance = nd (< }£ d) 
from the middle of joint (d = depth of joint.) Then we 
have the following relations : 

p (the mean press, per. sq. in.),p m (max. press, persq. in.), 
and p n (least press, per sq. in.) are proportional to the lines 
h (mid. width), a (max. base), and c (min. base) respectively, 
of a trapezoid, Fig. 400, (b), through whose centre of gravity 
P acts. But (§ 26) 



7 d a—c • 
na=-^ i.e., n 



6 — , — or a=h (6w+l) 



6 a-\-c h 

p m =p (6n-\-T). Hence the following table : 



If nd= }i d 
press. p m = 2 






% 8 d 



then the max. 

times the mean pressure. 



Case II. Let P fall outside the mid. third, a distance = 
nd (> }i d) from the middle of joint. Here, since the 
joint is not considered capable of withstanding tension, 
we have a triangle, instead of a trapezoid. Fig. 401. First 
compute the mean press, per sq. in. 

p — — — n s :„ '{ — , — or from this table : (lamina one 
(l—2n) 18 d inches 

foot thick). 



ARCHES OF MASONRY. 



437 



For nd= 4 / 18 d I 5 / 18 d 

P = 1/10 Pd\y 8 Pd 



6 As d 7 / 18 d 
VePd 



%Pd 



Visd 
%Pd 



/isd 

infinity. 



(d in inches.) 



Then the maximum pressure (at A, Fig. 401) p m , 
becomes known, in lbs. per sq. in. 



2? 



438 MECHANICS OF ENGINEERING. 



CHAPTEK XL 



ARCH-RIBS. 



364. Definitions and Assumptions. — An arch-rib (or elastic- 
arch, as distinguished from a block-work arch) is a rigid 
curved beam, either solid, or built up of pieces like a 
truss (and then called a braced arch) the stresses in which, 
under a given loading and with prescribed mode of sup- 
port it is here proposed to determine. The rib is sup- 
posed symmetrical about a vertical plane containing its 
axis or middle line, and the Moment of Inertia of any cross 
section is understood to be referred to a gravity axis of 
the section, which (the axis) is perpendicular to the said 
vertical plane. It is assumed that in its strained condi- 
tion under a load, the shape of the rib differs so little 
from its form when unstrained that the change in the ab- 
scissa or ordinate of any point in the rib axis (a curve) 
may be neglected when added (algebraically) to the co- 
ordinate itself ; also that the dimensions of a cross-section 
are small compared with the radius of curvature at any 
part of the curved axis, and with the span. 

365. Mode of Support. — Either extremity of the rib may be 
hinged to its pier (which gives freedom to the end-tangent- 
line to turn in the vertical plane of the rib when a load is 
applied); or may he fixed, i.e., so built-in, or bolted rigid- 
ly to the pier, that the end-tangent-line is incapable of 
changing its direction when a load is applied. A hinge 
may be inserted anywhere along the rib, and of course 



ARCH RIBS. 



439 



destroys the rigidity, or resistance to bending at that 
point. (A hinge having its pin horizontal ~| to the axis of 
the rib is meant). Evidently no more than three snch 
hinges could be introduced along an arch- rib between two 
piers ; unless it is to be a hanging structure, acting as a 
suspension-cable. 

366. Arch Rib as a Free Body. — In considering the whole 
rib free it is convenient, for graphical treatment, that no 
section be conceived made at its extremities, if fixed ; hence 
in dealing with that mode of support the end of the rib 
will be considered as having a rigid prolongation reach- 
ing to a point vertically above or below the pier junction, 
an unknown distance from it, and there acted on by a force 
of such unknown amount and direction as to preserve the 
actual extremity of the rib and its tangent line in the same 
position and direction as they really are. As an illustra- 
tion of this Fig. 402 
shows free an arch rib. 
OXB, with its extremi- 
ties and B fixed in the 
piers, with no hinges, q' 
and bearing two 
loads P, and P 2 . The 
other forces of the sys- 
tem holding it in equi- 
librium are the horizontal and vertical components, of the 
pier reactions (H, V, H n1 and V n ), and in this case of fixed 
ends each of these two reactions is a single force not in- 
tersecting the end of the rib, but cutting the vertical 
through the end in some point F (on the left ; and in G on 
the right) at some vertical distance c, (or d), from the end. 
Hence the utility of these imaginary prolongations OQF, 
and BEG, the pier being supposed removed. Compare 
Figs. 348 and 350. 

The imaginary points, or hinges, F and G, will be called 
abutments being such for the special equilibrium polygon 




Fig. 402. 



440 MECHANICS OF ENGINEERING. 

(dotted line), while and B are the real ends of the curved 
beam, or rib. 

In this system of forces there are five unknowns, viz.: V, 
V ut H = H nf and the distances c and d. Their determina- 
tion by analysis, even if the rib is a circular arc, is ex- 
tremely intricate and tedious ; but by graphical statics 
(Prof. Eddy's method ; see § 350 for reference), it is com- 
paratively simple and direct and applies to any shape of 
rib, and is sufficiently accurate for practical purposes. 
This method consists of constructions leading to the loca- 
tion of the " special equilibrium polygon " and its force 
diagram. In case the rib is hinged to the piers, the re- 
actions of the latter act through these hinges, Fig. 403, 
i.e., the abutments of the special 
equilibrium polygon coincide with ^-*4-^r 

the ends of the rib and B, and for \~y^~X' V \ 
a given rib and load the unknown / / ! p » K x 
quantities are only three V, F n , and V/ \ 
H; (strictly there are four ; but IX ~r4 * "~*hJ! ^ 

= gives H n = H). The solution FlG . 403 . 

by analytics is possible only for ribs of simple algebraic 
curves and is long and cumbrous ; whereas Prof. Eddy's 
graphic method is comparatively brief and simple and is 
applicable to any shape of rib whatever. 

i 367. Utility of the Special Equilibrium Polygon and its force 
diagram. The use of locating these will now be illustrated 
[See § 332]. As proved in §§ 332 and 334 the compres- 
sion in each " rod " or segment of the " special equilibrium 
polygon" is the anti-stress resultant of the cross sections in 
the corresponding portion of the beam, rib, or other struc- 
ture, the value of this compression (in lbs. or tons) being 
measured by the length of the parallel ray in the force 
diagram. Suppose that in some way (to be explained sub- 
sequently) the special equilibrium polygon and its force 
diagram have been drawn for the arch-rib in Fig. 404 hav- 
ing fixed ends, and B, and no hinges ; required the elastic 
stresses in any cross-section of the rib as at m. Let the 



ARCH BIBS. 



441 




Fig. 404. 

scale of the force-diagram on the right be 200 lbs. to the 
inch, say, and that of the space -diagram (on the left) 30 ft. 
to the inch. 

The cross section m lies in a portion TK, of the rib, cor- 
responding to the rod or segment be of the equilibrium 
polygon; hence its anti-stress-resultant is a force R 2 acting 
in the line be, and of an amount given in the force-diagram. 
Now B 2 is the resultant of V, H, and P lt which with the 
elastic forces at m form a system in equilibrium, shown in 
Fig. 405 ; the portion FO Tm being considered free. Hence 




Fig. 405. 



Fio. 406. 



taking the tangent line and the normal at m as axes we 
should have I (tang, comps.) = ; I (norm, comps.) = \ 
and 2 (moms, about gravity axis of the section at m) = 0, 
and could thus find the unknowns p lf p 2 , and J, which ap- 
pear in the expressions p Y F the thrust, &: the moment of 



442 MECHANICS OF ENGINEERING. 

the stress-couple, and J the shear. These elastic stresses 
are classified as in § 295, which see. p t and p 2 are lbs. per 
square inch, J is lbs., e is the distance from the horizontal 
gravity axis of the section to the outermost element of 
area, (where the compression or tension is p 2 lbs. per sq. 
in., as due to the stress-couple alone) while 1 is the " mo- 
ment of inertia " of the section about that gravity axis. 
[See §§ 247 and 295 ; also § 85]. Graphics, however, gives 
us a more direct method, as follows : Since R 2i in the line 
be, is the equivalent of V, H, and P lf the stresses at m will 
be just the same as if R 2 acted directly upon a lateral pro- 
longation of the rib at T (to intersect foFig. 405) as shown 
in Fig. 406, this prolongation Tb taking the place of TOF 
in Fig. 405. The force diagram is also reproduced here. 
Let a denote the length of the ~] from m's gravity axis 
upon be, and z the vertical intercept between m and be. 
For this imaginary free body, we have, 

from - (tang. compons.)=0, R, cos a—p x F 

and from J (norm. compons.)=0, R 2 sin a— J 

while from I (moms, about) ) -. r> p 2 I 

,, ., • i? \ n f we have Jx^a= ^-=- . 

the gravity axis oi m)=0, j e 

But from the two similar triangles (shaded ; one of them 
is in force diagram) a : z : : H :R 2 .'. R 2 a— Hz, whence we 
may rewrite these relations as follows (with a general state- 
ment), viz.: 

If the Special Equilibrium Polygon and Its Force Diagram Have 
Been Drawn for a given arch-rib, of given mode of support, 
and under a given loading, then in any cross-section of the 
rib, we have (F = area of section): 

The projection of the proper 
ray (of the force diagram) up- 
Un the tangent line of the rib 
drawn at the given section. 



ARCH RIBS. 443 



(2.) The Shear, i.e., J, = 
(upon which depends the 
shearing stress in the 
web). (See §§ 253 and 

256). 



The projection of the proper 

ray (of the force diagram) up- 
on the normal to the rib curve 
at the given section. 



(3.) The Moment of the 
stress couple, i.e., ^L , = H 



The product (Hz) of the H 
(or pole-distance) of the force - 
diagram by the vertical dis- 
tance of the gravity axis of the 
section from the spec, equilib- 
rium polygon. 
By the " proper ray " is meant that ray which is parallel 
to the segment (of the equil. polygon) immediately under 
or above which the given section is situated. Thus in 
Fig. 404, the proper ray for any section on TK is B 2 \ on 
KB, i? 3 ; on TO, R x . The projection of a ray upon any 
given tangent or normal, is easily found by drawing through 
each end of the ray a line "I to the tangent (or normal) ; 
the length between these 1 's on the tangent (or normal) is 
the force required (by the scale of the force diagram). We 
may thus construct a shear diagram, and a thrust diagram 
for a given case, while the successive vertical intercepts 
between the rib and special equilibrium polygon form a 
moment diagram. For example of the z of a point m is }4 
inch in a space diagram drawn to a scale of 20 feet to the 
inch, while H measures 2.1 inches in a force diagram con- 
structed on a scale of ten tons to the inch, we have, for the 
moment of the stress-couple at m, M=Hz= [2.1x10] tons 
X[^x20] ft. =210 ft. tons. 

368. — It is thus seen how a location of the special equili- 
brium polygon, and the lines of the corresponding force- 
diagram, lead directly to a knowledge of the stresses in all 
the cross-sections of the curved beam under consideration, 
bearing a given load ; or, vice versa, leads to a statement 
of conditions to be satisfied by the dimensions of the rib, 
for proper security. 

It is here supposed that the rib has sufficient lateral 



444 



MECHANICS OF ENGINEERING. 



bracing (with others which lie parallel with it) to prevent 
buckling sideways in any part like a long column. Before 
proceeding to the complete graphical analysis of the differ- 
ent cases of arch-ribs, it will be necessary to devote the 
next few paragraphs to developing a few analytical rela- 
tions in the theory of flexure of a curved beam, and to 
giving some processes in " graphical arithmetic." 

369. Change in the Angle Between Two Consecutive Rib Tan- 
gents when the rib is loaded, as compared with its value 
before loading. Consider any small portion (of an arch 
rib) included between two consecutive cross- sections ; Fig. 
407. KHGW is its unstrained form. Let EA, = ds, be 
the original length of this portion of the rib axis. The 
length of all the fibres (|j to rib-axis) was originally =ds 
(nearly) and the two consecutive tangent-lines, at E and A y 
made an angle = dd originally, with each other. While 
under strain, however, all the fibres are shortened equally 
an amount dX l9 by the uniformly distributed tangential 
thrust, but are unequally shortened (or lengthened, accord- 
ing as they are on one side or the other of the gravity axis 
E, or A, of the section) by the system of forces making 
what we call the " stress couple," among which the stress 
at the distance e from the gravity axis A of the section is 
called p 2 per square inch ; so that the tangent line at A' 
now takes the direction A'D "] to H'A'G' instead of A'C 
(we suppose the section at E to remain fixed, for conveni- 



t/ h /^ f (W* 




ARCH BIBS. 



445 



ence, since the change of angle between the two tangents 
depends on the stresses acting, and not on the new posi- 
tion in space, of this part of the rib), and hence the angle 
between the tangent-lines at E and A (originally = dd) is 
now increased by an amount CA'D — dip (or G'A'R = d<p); 
G'B.' is the new position of GH. We obtain the value of 
dcp as follows : That part (eW 2 ) of the shortening of the 
fibre at G, at distance e from A due to the force p/IF, is 

§ 201 eq. (1), dl 2 



rr\ Wo 

-^=-. But, geometrically, dX 2 also = 



Eed<p=p 2 ds 



(i-) 



But, letting M denote the moment of the stress-couple 
at section A (M depends on the loading, mode of support, 
etc., in any particular case) we know from § 295 eq. (6) that 

M=^~ t and hence by substitution in (1) we have 



d(p 



Mds 
EI 



(2) 



[If the arch-rib in question has less than three hinges, 
the equal shortening of the fibres due to the thrust (of 
the block in last figure) p Y F, will have an indirect effect on 
the angle dw. This will be considered later.] 

370. Total Change i.e. Cd<p in the Angle Between the End 

Tangents of a Rib, before and after loading. Take the ex- 
ample in Fig. 408 of a rib fixed at one end and hinged at 




Fig. 403. 



446 



MECHANICS OF ENGINEERING. 



the other. When the rib is unstrained (as it is supposed 
to be, on the left, its own weight being neglected; it is not 
supposed sprung into place, but is entirely without strain) 
then the angle between the end-tangents has some value 

d'= J dd= the sum of the successive small angles dd for 

each element ds of the rib curve (or axis). After loading, 
[on the right, Fig. 408], this angle has increased having, 
now a value 



0'+ r d<p, i.e., a value = d'-\- C -== 



a.) 



There must oe no hinge between 
I and B. 

% 371. Example of Equation (I) in Anal- 
p ysis. — A straight, homogeneous, pris- 
p matic beam, Fig. 409, its own weight 
* neglected, is fixed obliquely in a wall. 
After placing a load P on the free end, 
required the angle between the end- 
tangents. This was zero before load- 
ing .*. its value after loading is 




Mds 



=0+ ,'=o+ j 



By considering free a portion between and any ds of the 
beam, we find that M=Px=mom. of the stress couple. 
The flexure is so slight that the angle between any ds and 
its dx is still practically —a (§ 364), and .\ ds—dx sec «.. 
Hence, by substitution in eq. (I.) we have 



9 >=± rMds=l^rxdx-- 

* EI Jo EI Jo 



P sec a 
EI 



r-l cos a 
X 2 

Loo ' 



: P ( C(>sg ^ [Compare with § 237], 
2EI L r 



ARCH EIBS. 



44', 



It is now apparent that if both ends of an arch rib are 
fixed, when unstrained, and the rib be then loaded (within 
elastic limit, and deformation slight) we must have 

B 

J (Mds-t-EI) = zero, since f'=0. 

372. Projections of the Displacement of any Point of a Loaded 
Rib Relatively to Another Point and the Tangent Line at the Lat- 
ter. — (There must be no hinge between and B). Let 
be the point whose displacement is considered and B the 
other point. Fig. 410. If i?'s tangent-line is fixed while 
the extremity O-is not supported in any way (Fig. 410) 
then a load P put on, is displaced to a new position O n . 



vr ~r 




Fig. 410. 



Fig. 411. 



Fig. 412. 



"With as an origin and OB as the axis of X, the projec- 
tion of the displacement OO n upon X, will be called Ax, 
that upon Y, Ay. 

In the case in Fig. 410, O's displacement with respect to 
B and its tangent-line BT, is also its absolute displacement 
in space, since neither B nor BT has moved as the rib 
changes form under the load. In Fig. 411, however, the 
extremities and B are both hinged to piers, or supports, 
the dotted line showing its form when deformed under a 
load. The hinges are supposed immovable, the rib being 
free to turn about them without friction. The dotted line 
is the changed form under a load, and the absolute dis- 
placement of is zero ; but not so its displacement rela- 
tively to B and B's tangent BT, for BT has moved to a 
new position BT'. To find this relative displacement con- 
ceive the new curve of the rib superposed on the old in 
a way that B and BT may coincide with their original po- 



448 



MECHANICS OF ENGINEERING. 



sitions, Fig. 412. It is now seen that O's displacement 
relatively to B and B T is not zero but = n , and has a 
small Ax bnt a comparatively large Ay. In fact for this 
case of hinged ends, piers immovable, rib continuous be- 
tweer them, and deformation slight, we shall write Ax — 
zero as compared with Ay, the axis X passing through OB). 

373. Values of the X and Y Projections of O's Displacement Rela- 
tively to B and B's Tangent ; the origin being taken at 0, 



f-^8 



^/V G . 




Fig. 413. Let the co- 
ordinates of the dif- 
ferent points E, D, G, 
etc., of the rib, re- 
ferred to and an 
arbitrary X axis, be 
x and y, their radial 
distances from be- 
ing u (i.e., u for G, u f 
for D, etc.; in gener- 
al, u). OEDG is the jl i ^ 

unstrained form of the i * j 

rib, (e.g., the form it Fl ' G . 413 

would assume if it lay flat on its side on a level platform, 
under no straining forces), while O n E"D'GB is its form 
under some loading, i.e., under strain. (The superposi- 
tion above mentioned (§ 372) is supposed already made if 
necessary, so that BT is tangent at B to both forms). 
Now conceive the rib OB to pass into its strained condi- 
tion by the successive bending of each ds in turn. The 
straining or bending of the first ds, BG, through the small 
angle dip (dependent on the moment of the stress couple 
at G in the strained condition) causes the whole finite piece 
00 to turn about G as a centre through the same small 
angle dip ; hence the point describes a small linear arc 
00'=dv, whose radius = u the hypothenuse of the x and 
y of O, and whose value .*. is bv—udip. 

Next let the section 1), now at D\ turn through its 
proper angle dip' (dependent on its stress-couple) carrying 



ARCH RIBS. 449 

with it the portion D f 0\ into the position D'0" f making 
0' describe a linear arc 0' 0" =(dv)' =u'd<p' f in which u'= 
the hypothenuse on the x' and y f (of D), (the deformation 
is so slight that the co-ordinates of the different points 
referred to and X are not appreciably affected). Thus, 
each section having been allowed to turn through the an- 
gle proper to it, finally reaches its position, n , of dis- 
placement. Each successive dv, or linear arc described by 
0, has a shorter radius. Let dx, (dx)', etc., represent the 
projections of the successive (8v)'a upon the axis X; and 
similarly dy, (%)' etc., upon the axis Y. * Then the total X 
projection of the curved line . . . . a will be 

Jx= I dx and similarly Ay — joy . . . (1) 

But d v = u d <p, and from similar right -triangles, 
S x: dv : : y : u and dy : dv : : x : u .: dx = yd<p and dy=xd<p ; 
whence, (see (1) and (2) of §369) 

Myds 



Jx = f 8x = fyd 9 = £-^p ■ . . (II.) 



and 



Mxds 



Ay=f d y=f xdtp =f Q ^f. . . (HI.) 

If the rib is homogeneous E'is constant, and if it is of 
constant cross-section, all sections being similarly cut by 
the vertical plane of the rib's axis (i.e., if it is a " curved 
prism ") J, the moment of inertia is also constant. 

374. Recapitulation of Analytical Relations, for reference* 
(Not applicable if there is a hinge betiveen and B) 

Total Change in Angle between ) _ r^Mds fJ . 

tangent-lines and B j J ~E£ • • • • K 1 -) 

The X-Projection of O's Displacement "1 

Relatively to B.and B's tangent- Bj[f ^ 

line ; (the origin being at 0) I — I ty T • " ■ \ ±L -) 

and the axes I and 7 1 to | Jo hi 

each other) I 



450 



MECHANICS OF ENGINEERING. 



The Y-Projection of 0's Displacement, 
etc., as above. 



\-r- 



Mxds 



EI 



(III.) 



Plere x and y are the co-ordinates of points in the rib- 
curve, ds an element of that curve, M the moment of the 
stress-couple in the corresponding section as induced by 
the loading, or constraint, of the rib. 

(The results already derived for deflections, slopes, etc., 
for straight beams, could also be obtained from these 
formulae, I., II. and III. In these formulae also it must 
be remembered that no account has been taken of the 
shortening of the rib-axis by the thrust, nor of the effect 
of a change of temperature.) 

374a. Resumfe of the Properties of Equilibrium Polygons and 
their Force Diagrams, for Systems of Vertical Loads. — See §§ 335 
to 343. Given a system of loads or vertical forces, P l9 P 2i 

etc., Fig. 414, and 
two abutment verti- 
cals, F' and G' \ if 
% we lay off, vertically, 
to form a " load- 
line," 1 . . 2 = P l3 2. . . 
3=P 2 > etc., select any 
Pole, lf and join O x 
... 1, X . . . 2, etc. ; 
also, beginning at 
any point F x in the 
. . 1 to intersect the 




0, 



Fig. 414. 

vertical F' } if we draw F x . . . a || to 
line of Pj ; then ab || to X . . 2, and so on until finally a 
point 6rj, in G\ is determined; then the figure F lt abc G x is 
an equilibrium polygon for the given loads and load verti- 
cals, and l . . . 1234 is its " force diagram." The former 
is so called because the short segments F x a ab, etc., if 
considered to be rigid and imponderable rods, in a vertical 
plane, hinged to each other and the terminal ones to abut- 
ments F x and G u would be in equilibrium under the given 
loads hung at the joints. An infinite number of equilib- 



ARCH-BIBS. 451 

riuni polygons may be drawn for the given loads and 
abutment-verticals, by choosing different poles in the force 
diagram. [One other is shown in the figure ; 2 is its 
pole. \Fi G Y and F 2 G 2 are abutment lines.)] For all of 
these the following statements are true : 

(1.) A line through the pole, || to the abutment line cuts 
the load-line in the same point n', whichever equilibrium 
polygon be used ( .• any one will serve to determine n% 

(2.) If a vertical CD be drawn, giving an intercept z' in 
each of the equilibrium polygons, the product Hz' is the 
same for all the equilibrium polygons. That is, (see Fig. 
414) for any two of the polygons we have 

H X :H 2 :: z/ : a/ ; or H 2 z 2 ' = H, «/. 

(3.) The compression in each rod is given by that 
" ray " (in the force diagram) to which it is parallel. 

(4.) The " pole distance " H } or "] let fall from the pole 
upon the load-line, divides it into two parts which are the 
vertical components of the compressions in the abutment- 
rods respectively ( the other component being horizontal) ; 
H is the horizontal component of each (and, in fact, of 
each of the compressions in all the .other rods). The 
compressions in the extreme rods may also be called the 
abutment reactions (oblique) and are given by the extreme 
rays. 

(5.) Three Points [not all in the same segment (or rod)] 
determine an equilibrium polygon for given loads. Hav- 
ing given, then, three points, we may draw the equilibrium 
polygon by §341. 

375. Summation of Products. Before proceeding to treat 
graphically any case of arch-ribs, a few processes in 
graphical arithmetic, as it may be called, must be pre- 
sented, and thus established for future use. 

To make a summation of products of two factors in each 
by means of an equilibrium polygon. 



452 



MECHANICS OF ENGINEERING. 



Construction. Suppose it required to make the summa- 
tion I (x z) i. e., to sum the series 



x 1 z 1 +x 2 z 2 -\-x s z z + 



by graphics. 





K 






/ 




i 






. 


i K 
of t 


1 


f 


V 






S^ • ! 


'Z 


. 


c 2 . 


> 


r 


2' 1 /is. 


r "5 




<r"7 


T 




s s^ 


^■"-SfcL 


u 


Z 






■2 1 


a 1 


F 


*-a: T * 


h 




1 i 








-X 


-» 


! | 






* 




-X; 


v i 






* 






■*ri—\ 




p 


« 






—-■^xirf—*'. 




* 




— 


-Zs-i 1 


■ >» 




Having first arranged the terms in the order of magni- 
tude of the x's, we proceed as follows : Supposing, for 
illustration, that two of the s's (% and z 4 ) are negative 
(dotted in figure) see Fig. 415. These quantities x and z 
may be of any nature whatever, anything capable of being 
represented by a length, laid off to scale. 

First, in Fig. 
416, lay off the 
s's in their 
order, end to 
end, on a ver- 
tical load-line 
taking care to 
lay off z % and 
3 4 upiuard in 
their turn. 
Take any con- 
fig. 4i6. venient pole 

1, ... 2, etc.; then, having pre- 
viously drawn vertical lines whose horizontal distances 
from an extreme left-hand vertical F' are made = x l} x iy 
a? 3 , etc., respectively, we begin at any point F, in the verti- 
cal F', and draw a line || to ... 1 to intersect the x x ver- 
tical in some point ; then 1' 2' II to . . . 2, and so on, fol- 
lowing carefully the proper order. Produce the last seg- 
ment (6' ... G in this case) to intersect the vertical F' in 
some point K. Let KF =k (measured on the same scale 
as the x's), then the summation required is 

2Y 3 (xz) = Hk. 

H is measured on the scale of the a's, which need not be 
the same as that of the a?'s ; in fact the z*s may not be the 
same kind of quantity as the x's. 

[Proof. — From similar triangles H:z l ::x L : h l9 .*. x l z i =Hh i \ 
and " " " H: z 9 :: x 2 : k 2i .*. x 2 z 2 =Bk 2 . 



Fig. 415. 

; draw the rays 



ABCH-RIBS. 



45,1 



and so on. But H (k l +k 2 -\-etc.)=B:xFK=Hk]. 

376. Gravity Vertical. — From the same construction in 
Fig. 415 we can determine the line of action (or gravity 
vertical) of the resultant of the parallel vertical forces z lf 
z 2 , etc. (or loads); by prolonging the first and last segments 

to their intersection at 
0. The resultant of the 
system of forces or loads 
acts through C and is 
vertical in this case ; its 
value being = I (z), 
that is, it = the length 
1 ... 7 in the force dia- 
gram, interpreted by the 
proper scale. It is now 
supposed that the z's 
represent forces, the #'s 
being their respective 
lever arms about F. If 
the z's represent the 
areas of small finite por- 
tions of a large plane 
figure, we may find a 
gravity-line (through C) 
of that figure by the 
above construction; each 
z being-applied through 
the centre of gravity of 
its own portion. 

Calling the distance 

x between the verticals 

through C and F, we 

k 3 have also x . I (z) = 

I (xz) because I (z) is 

the resultant of the || z's. 

ki This is also evident from 

the proportion (similar 

triangles) 

H : (1 . . 7) :: x : h 




454 MECHANICS OF ENGINEERING. 

376 a. Moment of Inertia (of Plane Figure) by Graphics.— Fig. 
416 a. I N = ? First, for the portion on right. Divide OB 
mto equal parts each = Ax. Let z u z 2} etc., be the middle 
ordinates of the strips thus obtained, and x ly etc. their 
abscissas (of middle points). 

Then we have approximately 



7 N for OB=Ax.z L x L 2 -{-4x.z 2 x 2 2 -\- 

=Jx[(z l x 1 )x l -\-(z 2 x 2 )x 2 + ...].. (1) 

But by §375 we may construct the products z l x 1 ,z 2 x 2 , etc., 
taking a convenient H', (see Fig. 416, (&)), and obtain k lt k 2 , 
etc., such that ZyX x = H'k^ z 2 x 2 = H'k 2 , etc. Hence eq. (1) 
becomes : 

i N for OB approx.=i?'z/x[& 1 a; 1 +&2^2+ ...]... (2) 

By a second use of § 375 (see Fig. 416 c) we construct I, 
such that A^ + k 2 x 2 + ....= H"l [H" taken at con- 
venience]. .*. from eq. (2) we have finally, (approx.), 

I N for OB=B'H"lAx (3) 

For example if OB = 4 in., with four strips, Ax would = 
1 in.; and if H f = 2 in., B" = 2 in., and I = 5.2 in., then 

I N for OB = 2x2x5.2x1.0=20.8 biquad. inches. 

The I s for OX, on the left of N, is found in a similar 
manner and added to I N for OB to obtain the total 7~ N . The 
position of a gravity axis is easily found by cutting the 
shape out of sheet metal and balancing on a knife edge ; or 
may be obtained graphically by § 336 ; or 376. 

377. Construction for locating a line vm (Fig. 417) at (a), in 
the polygon FG in such a position as to satisfy the two 
following conditions with reference to the vertical inter- 
cepts at 1, 2, 3, 4, and 5, between it and the given points 
1, 2, 3, etc., of the perimeter of the polygon. 



ARCH-RIBS. 



455 



Condition I. — (Calling these intercepts u lf u 2 , etc., and their 
horizontal distances from a given vertical F, x l} x. 2i etc.) 

2 (u) is to = ; i.e., the sum of the positive ^'s must be 
numerically = that of the negative (which here are at 1 
and 5). An infinite number of positions of vm will satisfy 
condition I. 



(a) 



Condition II. — I (ux) is to = ; i.e., the sum of the 

m moments of the positive w's 
about F must = that of the 

G ( a ) negative u's. i.e., the moment 
of the resultant of the posi- 

m tive us must = that of the 
resultant of the negative ; 

G and .*. (Condit. I being 

m already satisfied) these two 
resultants must be directly 

g{ c ) opposed and equal. But the 
ordinates u in (a) are indi- 

m vidually equal to the differ- 
ence of the full and dotted 
ordinates in (b) with the 
same x's .*. the conditions 
may be rewritten : 

I. I (full ords. in (b))= 
I (dotted ords. in (&)) 

II. I [each full ord. in (b) 
X its x~\ = I [each dotted 
ord. in (b) X its x] i.e., the 

fig. 417. centres of gravity of the full 

and of the dotted in (b) must lie in the same vertical. 

Again, by joining v G, we may divide the dotted ordi- 
nates of (b) into two sets which are dotted, and broken, re- 
spectively, in (c) Then, finally, drawing in (d), 

B, the resultant of full ords. of (c) 
T, " " " broken " " " 

T, " " " dotted " " " 




456 MECHANICS OF ENGINEERING. 

we are prepared to state in still another and final form the 
conditions which vm must fulfil, viz. : 

(I.) T+T must = R\ and (II.) The resultant of T 
and T' must act in the same vertical as R. 

In short, the quantities T, T', and R must form a bal- 
anced system, considered as forces. All of which amounts 
practically to this : that if the verticals in which T and T' 
act are known and R be conceived as a load supported by 
a horizontal beam (see foot of Fig. 417, last figure) resting 
on piers in those verticals, then T and T' are the respec- 
tive reactions of those piers. It will now be shown that the 
verticals of T and T' are easily found, being independent of 
the position of vm ; and that both the vertical and the mag- 
nitude of R, being likewise independent of vm, are deter- 
mined with facility in advance. For, if v be shifted up 
or down, all the broken ordinates in (c) or (d) will change 
in the same proportion (viz. as vF changes), while the 
dotted ordinates, though shifted along their verticals, do 
not change in value ; hence the shifting of v affects neither 
the vertical nor the value of T', nor the vertical of T. 
The value of T, however, is proportional to vF. Similar- 
ly, if m be shifted, up or down, T' will vary proportionally 
to mG, but its vertical, or line of action, remains the same. 
T is unaffected in any way by the shifting of m. R, de- 
pending for its value and position on the full ordinates of 
(c) Fig. 417, is independent of the location of vm. We 
may .*. proceed as follows : 

1st. Determine R graphically, in amount and position, 
by means of § 376. 

2ndly. Determine the verticals of T and T f by any trial 
position of vm (call it v 2 m 2 ), and the corresponding trial 
values of T and T' (call them T 2 and T' 2 ). 

3rdly. By the fiction of the horizontal beam, construct 
(§ 329) or compute the true values of T and T\ and then 
determine the true distances vF and mG by the propor- 
tions 

vF :v 2 F:: T : T 2 and mG : m 2 G : : T : T\. 



AECH-RIBS. 



457 



Example of this. Fig. 418. (See Fig. 417 for s and t) 

From A toward B in (e) Fig. 418, lay off the lengths (or 
lines proportional 
to them) of the full 
ordinates 1, 2, etc., 
of (/). Take any 
pole U and draw the 
equilibrium poly- 
gon of (/) and pro- 
long its extreme seg- 
ments to find C and 
thus determine i?'s 
vertical. B is repre- 
sented by AB. In 
(g) [same as (/) but 
shifted to avoid 
complexity of lines] 
draw a trial v 2 m 2 and 
join v 2 G 2 . Deter- 
mine the sum T 2 of 
the broken ordi- IFig. 418. 

nates (between v 2 G 2 ana F 2 G 2 ) and its vertical line of ap- 
plication, precisely as in dealing with R ; also T 2 that of 
the dotted ordinates (five) and its vertical. Now the true 
T=Rt+(s+t) and the true T'=Bs+(s+t). Hen ce com- 
pute vF=(T-r-T 2 ) v 2 F 2 and ^G=(T'+T' 2 ) ~m~G 2 , and by 
laying them off vertically upward from F and G respec- 
tively we determine v and m, i.e., the line vm to fulfil the 
conditions imposed at the beginning of this article, rela- 
ting to the vertical ordinates intercepted between vm and 
given points on the perimeter of a polygon or curve. 

Note (a). If the verticals in which the intercepts lie are 
equidistant and quite numerous, then the lines of action 
of T 2 and T' 2 will divide the horizontal distance between 
F and G into three equal parts. This will be exactly true 
in the application of this construction to § 390. 

Note (b). Also, if the verticals are symmetrically placed 
about a vertical line, (as will usually be the case) v 2 m 2 is 




458 



MECHANICS OF ENGINEERING. 



best drawn parallel to FG, for then T 2 and T\ will be 
equal and equi-distant from said vertical line. 

378. Classification of Arch-Ribs, or Elastic Arches, according 
to continuity and modes of support. In the accompany- 
ing figures the full curves show the unstrained form of the 
rib (before any load, even its own weight, is permitted to 
come upon it) ;the dotted curve shows its shape (much ex- 
aggerated) when bearing a load. For a given loading 
Three Conditions must be given to determine the special 
equilibrium polygon (§§ 366 and 367). 

Class A. — Continuous rib, free to slip laterally on the 
piers, which have smooth horizontal surfaces, Fig. 420. 

This is chiefly of theoretic interest, its consideration 
being therefore omitted. The pier reactions are neces- 
sarily vertical, just as if it were a straight horizontal 
beam. 

Class B. Rib of Three Hinges, two at the piers and one 
intermediate (usually at the crown) Fig. 421. Fig. 36 also 
is an example of this. That is, the rib is discontinuous 
and of two segments. Since at each hinge the moment oi 
the stress couple must be be zero, the special equilibrium 
polygon must pass through the hinges. Hence as three 
points fully determine an equilibrium polygon for given 
load, the special equilibrium is drawn by § 341. 




Fig. 420. 



Fig. 421. 



[§ 378a will contain a construction for arch-ribs of three 
hinges, when the forces are not all vertical.] 

Class C. Rib of Two Hinges, these being at the piers, the 
rib continuous between. The piers are considered im- 
movable, i.e., the span cannot change as a consequence of 
loading. It is also considered that the rib is fitted to its 



AKCH IUBS. 



459 



hinges at a definite temperature, and is then under no con- 
straint from the piers (as if it lay flat on the ground), not 
even its own weight being permitted to act when it is fi- 
nally put into position. When the "false works" 
or temporary supports are removed, stresses are in- 
duced in the rib both by its loading, including its 
own weight, and by a change of temperature. Stresses 
due to temperature may be ascertained separately and 
then combined with those due to the loading. [Classes 
A and B are not subject to temperature stresses.] Fig. 

422 shows a rib of two hinges, 
at ends. Conceive the dotted 
curve (form and position un- 
der strain) to be superposed 
on the continuous curve 
(form before strain) in such 
a way that B and its tangent 
line (which has been dis- 
placed from its original position) may occupy their pre- 
vious position. This gives us the broken curve n B. 00 a 
is .*. O's displacement relatively to B and B's tangent. 
Now the piers being immovable n B (right line) =0i? ; i.e., 
the X projection (or Ax) of 00 a upon OB (taken as an axis 
of X) is zero compared with its Ay. Hence as one condi- 
tion to fix the special equilibrium polygon for a given load- 
ing we have (from § 373) 




r B [Myds+EI]=0 



(1) 



The other two are that the ) must pass through 



\ 



(2) 
special equilibrium polygon ) " " " B . (3) 

Class D. Eib with Fixed Ends and no hinges, i.e., continu- 
ous. Piers immovable. The ends may be fixed by being 
inserted, or built, in the masonry, or by being fastened to 
large plates which are bolted to the piers. [The St. Louis 
Bridge and that at Coblenz over the Bhine are of this 
class.] Fig. 423. In this class there being no hinges we 



460 



MECHANICS OF ENGINEERING. 




have no point given in advance through which the special 
equilibrium polygon must pass. However, since O's dis- 
placement relatively (and absolutely) to B and B's tangent 
is zero, both Ax and Ay [see § 373] = zero. Also the tan- 
gent-lines both at and B being 
fixed in direction, the angle be- 
tween them is the same under 
loading, or change of temperature, 
as when the rib was first placed 
in position under no strain and at 
a definite temperature. 
Hence the conditions for locating the special equilibrium 
polygon are 

r^ Mds _ . s> Myds _ Q . f Mxds _ a 
Jo EI ' Jo ~~E1~ ' Jo EI 

In the figure the imaginary rigid prolongations at the 
ends are shown [see § 366]. 

Other designs than those mentioned are practicable 
(such as : one end fixed, the other hinged ; both ends fixed 
and one hinge between, etc.), but are of unusual occur- 
rence. 



Fig. 423. 



378a. Rib of Three Hinges. Forces not all Vertical. If the 
given rib of three hinges upholds a roof, the wind-press- 
ure on which is to be considered as well as the weights of 
the materials composing the roof-covering, the forces will 
not all be vertical. To draw the special equil. polygon in 

such a case the following 
construction holds : Re- 
quired to draw an equilib- 
rium polygon, for any 
plane system of forces, 
through three arbitrary 
points, A, p and B ; Fig. 
B423a. Find the line of 
action of B l3 the resultant 
of all the forces occurring 
between A and p; also,, 




Fig. 423a. 






ARCH-KIBS. 



461 



that of R 2} the resultant of all forces between [p and B ; 
also the line of action of R, the resultant of R i and R 2) [see 
§ 328.] Join any point ill in R with A and also with B, 
and join the intersections N and 0. Then A N will be the 
direction of the first segment, B that of the last, and 
NO itself is the segment corresponding to p (in the der 
sired polygon) of an equilibrium polygon for the given 
forces. See § 328. If A N'p 0' B are the corresponding 
segments (as yet unknown) of the desired equil. polygon, 
we note that the two triangles MNO and M'N 0\ having 
their vertices on three lines which meet in a point [i.e., R 
meets i? x and R 2 in (7'], are homological [see Prop. VII. of 
Introduc. to Modern Geometry, in Chauvenet's Geometry,] 
and that . ■ . the three intersections of their corresponding 
sides must lie on the same straight line. Of those inter-: 
sections we already have A and B, while the third must be 
at (7, found at the intersection of AB and NO. Hence by 
connecting C and p, we determine N' and 0'. Joining 
NA and O'B, the first ray of the required force diagram will 
be || to NA, while the last ray will be || to O'B, and thus 
the pole of that diagram can easily be found and the cor- 
responding equilibrium polygon, beginning at A, will pass 
through p and B. 

(This general case includes those of §§ 341 and 342.) 

379. Arch-Rib of two Hinges; by Prof. Eddy's Method.* 
[It is understood that the hinges are at the ends.] Re- 
quired the location of the special equilibrium polygon. We 
here suppose the rib homogeneous (i.e., the modulus of 
Elasticity E is the same throughout), that it is a " curved 
prism " (i.e., that the moment of inertia I of the cross- 
section is constant), that the piers are on a level, and that 
the rib-curve is symmetrical about a vertical line. Fig. 

424. For each point m of the rib 
curve we have an x and y (both 
known, being the co-ordinates of 
the point), and also a z (intercept 
t- between rib and special equilib- 
^ rium polygon) and a z' (intercept 




* P. 25 of Prof. Eddy's book ; see reference in preface of this work. 



4G2 



MECHANICS OF ENGENEEKING. 



between the spec. eq. pol. and the axis X (which is OB), 
The first condition given in § 378 for Class C may be 
transformed as follows, remembering [§ 367 eq. (3)] that 
M — Hz at any point m of the rib (and that EI is con- 
stant). 

*. £ Myds = 0, i.e., .g j\yds = . • . j\yds = 



EI 

but 
z =y - 



}B B B 

• ' • I (y — z')yds=0; i.e., f yyds = f yz'ds . (1) 



In practical graphics we can not deal with infinitesimals ; 
hence we must substitute As a small finite portion of the 
rib-curve for ds; eq. (1) now reads 2* yy As = 2' B yz' As. 
But if we take all the As's equal, As is a common factor 
and cancels out, leaving as a final form for eq. (1) 

Z \yy) = I*{yz') . . . (1)' 

The other two conditions are that the special equilibrium 
polygon begins at and ends at B. (The subdivision of 
the rib-curve into an even number of equal As's will be ob- 
served in all problems henceforth.) 

379a. Detail of the Construction. Given the arch-rib B> 
Fig. 425, with specified loading. Divide the curve into 



12 3 4 5 G 7 8 

ii I I I lii 

)A i PK | 

i/l f 3 \ h I \| 




Fig. 42. 




ARCH RIBS. 



463 



eight equal ^s's and draw a vertical through the middle 
of each. Let the loads borne by the respective Js's be 
F lt P 2 , etc., and with them form a vertical load-line A C to 
some convenient scale. With any convenient pole 0" 
draw a trial force diagram 0" AC, and a corresponding 
trial equilibrium polygon F G, beginning at any point in 
the vertical F. Its ordinates %", 2 2 ", etc., are propor- 
tional to those of the special equil. pol. sought (whose 
abutment line is OB) [§ 374a (2)]. We next use it to de- 
termine n' [see § 374a]. We know that OB is the " abut- 
ment-line " of the required special polygon, and that . * . 
its pole must lie on a horizontal through n'. It remains 
to determine its H, or pole distance, by equation (1)' just 
given, viz. : 2? yy = 2?yz'. First by § 375 find the value 
of the summation 2f(yy), which, from symmetry, we may 
write = 22 1 %yy)=2[y 1 y l +y& i +y& i +y#J 

y x Hence, Fig. 426, we obtain 

2\ (yy)=2[H k] 

Next, also by § 375, see Fig. 
427, using the same pole dis- 
tance H as in Fig. 426, we 
find 

2\(yz")=H < A"\ i.e., 




2/i*i" +2/2*2' 
H n h". 



+ 2/3%" +y&" = 



2\(yz") = H (k"+k r 



Again, since 2\ (yz")= y 8 z 8 " 

+ 2/7*7" + 2/6*6" + 2/5*5" which 
from symmetry (of rib) 

= 2/l*8" + 2/2* 7 "+2/3*6" + 2/4*5", 

we obtain, Fig. 428, 

Il(2/*'') = #o&r",(same H ); 

and .-. 

If now we find that k i "+k t "=2h 



464 MECHANICS OE ENGINEERING. 

the condition 2f (yy) = 2\ (yz") is satisfied, and the pole 
distance of our trial polygon in Fig. 425, is also that of 
the special polygon sought; i.e., the z" 's.are identical in 
value with the z"s of Fig. 424. In general, of course, we 
do not find that k"-\-h" = 2k. Hence the z" 's must all 
be increased in the ratio 2k : (&/'+&/') to become equal to 
fche 2''s. That is, the pole distance H of the spec, equil- 
polygon must be 

jt_ k{'-\-k" jr,, (in which H" = the pole distance of the 
2k trial polygon) since from §339 the ordi- 

nates of two equilibrium polygons (for the same loads) 
are inversely as their pole distances. Having thus found 
the H of the special polygon, knowing that the pole must 
lie on the horizontal through n', Fig. 425, it is easily 
drawn, beginning at 0. As a check, it should pass through 
B. 

For its utility see § 367, but it is to be remembered that 
the stresses as thus found in the different parts of the 
rib under a given loading, must afterwards be combined 
with those resulting from change of temperature and the 
shortening of the rib axis due to the tangential thrusts, 
before the actual stress can be declared in any part. 

[Note. — If the " moment of inertia," 7, of the rib-sec- 
tion is different at different sections, i.e., if /is variable, 

foreq.(l)'we may write ^ ((?{)- ™(y- *') . . . (1)" 

(where n = -, 1 Q being the moment of inertia of a particu- 
lar section taken as a standard and /that at any section 

of rib) and in Fig. 426, use the ^ of each As instead of y 

n 

a" 

in the vertical " load-line," and — for z" in Figs. 427 and 

n 

428]. 



AECH-EIBS. 



4G5 



380. Arch Rib of Fixed Ends and no Hinges,— Example of 
Class D. Prof. Eddy's Method.* As before, E and I are 
constant along the rib Piers immovable. Rib curve 
sym etrical about a vertical line. Fig. 429 shows such a 
rib under any loading Its span is OB, which is taken as 
an axis X. The co-ordinates of any point m' of the rib 
curve are x and y, and z is the vertical intercept between 
m' and the special equilibrium polygon (as yet unknown, 
but to be constructed). Prof. Eddy's method will now be 

given for finding tha spe- 
v » cial equil. polygon. The 
three conditions it 
must satisfy (see § 370, 
Class T>, remembering 
that E and /are constant 
and that M = Ih from 
§ 367) are 




Fig. 429. 



Czds=0 ; Cxzds= ;0 and Cyzds=0 . . (1) 

c/o c/o c/o 

Now suppose the auxiliary reference line (straight) vm 
to have been drawn satisfying the requirements, with 
respect to the rib curve that 



Cz'ds = ; and Cxz'ds = 

e/o c/o 



(2) 



in which z is the vertical distance of any point w! from 
vm and x the abscissa of m' from 0. 

From Fig. 429, letting z" denote the vertical intercept 
(corresponding to any m') between the spec, polygon and 
the auxiliary line vm, we have z=z'—z", hence the three 
conditions in (1.) become 



r{z'-z")ds=0; i.e., see eqs. (2) C. z"ds=0 



(3) 



* P. 14 of Prof. Eddv's book ; see reference in preface of this work. 



466 MECHANICS OF ENGINEERING. 

B B 

Cx (z r —z r )ds=0; i.e., see eqs. (2) C xz"ds=o (4) 

c/o c/ o 

- d X^-^ S=; °---poSnX^ = ^ 8 ''* • (5) 

provided vm has been located as prescribed. 

For graphical purposes, having subdivided the rib curve 
into an even number of small equal Js's, and drawn a verti- 
cal through the middle of each, we first, by § 377, locate 
vm to satisfy the conditions 

2?(«') =0 and 2?(a*')=0 . . (6) 

(see eq. (2) ; the As cancels out) ; and then locate the 
special equilibrium polygon, with vm as a reference-line, 
by making it satisfy the conditions. 

25(0=0 • ( 7 ); ^o>O=0 • (8); Z?(yz")=Z«(yz') . (9) 

(obtained from eqs. (3), (4), (5) by putting ds=As, and can- 
celling). 

Conditions (7) and (8) may be satisfied by an infinite 
number of polygons drawn to the given loading. Any one 
of these being drawn, as a trial polygon, we determine for it 
the value of the sum 2l(yz") by § 375, and compare it with 
the value of the sum I^(yz') which is independent of the 
special polygon and is obtained by § 375. [N.B. It must 
be understood that the quantities (lengths) x, y, z, z', and z", 
here dealt with are those pertaining to the verticals drawn 
through the middles of the respective ^s's, which must be 
sufficiently numerous to obtain a close result, and not to 
the verticals in which the loads act, necessarily, since these 
latter may be few or many according to circumstances, see 
Fig. 429]. If these sums are not equal, the pole distance 
of the trial equil. polygon must be altered in the proper 
ratio (and thus change the s'"s in the inverse ratio) neces- 
sary to make these sums equal and thus satisfy condition 
(9). The alteration of the s'"s, all in the same ratio, will 



AKCH-HLBS. 



467 



not interfere with conditions (7) and (8) which are already 
satisfied. 



381. Detail of Construction of Last Problem. Symmetrical Arch- 
Rib of Fixed Ends. — As an example take a span of the St. 
Louis Bridge (assuming /constant) with " live load" cov- 
ering the half span on the left, Eig. 430, where the vertical 




Ty=fc5tM=i4. 



Fig. 430. 



scale is much exaggerated for the sake of distinctness*. 
Divide into eight equal ds's. (In an actual example sixteen 
or twenty should be taken.) Draw a vertical through the 



* Each arch-rib of the St. Louis bridge is a built up or trussed rib of steel about 520 
ft. span and 52 ft rise, in the form of a segment of a circle . Its moment of inertia, 
however, is not strictly constant, the portions near each pier, of a length equal to one 
twelfth of the span, having a value of /one-half greater than that of the remainder of 
the arc. 



468 MECHANICS OF ENGLN T EEKING. 

middle of each As. P l} etc., are the loads coming upon the 
respective Js's. 

First, to locate vm, by eq. (6) ; from symmetry it must 
be horizontal. Draw a trial vm (not shown in the figure) 
and if the (+^')' s exceed the ( — «')'s by an amount z Q ', the 

true vm will lie a height — zj above the trial vm (or be- 

n 

low, if vice versa) ; n = the number of As's. 

Now lay off the load-line on the right, (to scale), 
take any convenient trial pole 0"' and draw a correspond- 
ing trial equil. polygon F"'G m . In F'"G"\ by § 377, 
locate a straight line v"m" so as to make Z*(z'")=Q and 
2*(aw"')=0 (see Note (6) of § 377). 

[We might now redraw F"'G'" in such a way as to bring 
v , "m"' into a horizontal position, thus : first determine a 
point n'" on the load-line by drawing 0"'n'" || to v'"m'" take 
a new pole on a horizontal through n'", with the same 
H 1 ', and draw a corresponding equil. polygon; in the lat- 
ter v'"m'" would be horizontal. We might also shift this 
new trial polygon upward so as to make v'"m' ' and vm 
coincide. It would satisfy conditions (7) and (8), having 
the same s""s as the first trial polygon ; but to satisfy con- 
dition (9) it must have its s""s altered in a certain ratio, 
which we must now find. But we can deal with the indi- 
vidual sT"s just as well in their present positions in Fig. 
430.] The points E and L in vm, vertically over E r " and 
L"' in v'"m!" , are now fixed ; they are the intersections of the 
special polygon required, with vm. 

The ordinates between v'"m f " and the trial equilibrium 
polygon have been called z'" instead of z"\ they are pro- 
portional to the respective 2"'s of the required special 
polygon. 

The next step is to find in what ratio the (z"')'s need to 
be altered (or H" altered in inverse ratio) in order to be- 
come the (z")'s ; i.e., in order to fulfil condition (9), viz.: 



AKCH-KLBS. 



409 




Z\{yz") = l\(yz>) . (9) 

This may be done pre- 
cisely as for the rib with 
two hinges, but the nega- 
tive (s'")'s must be prop- 
erly considered (§ 375) 
See Fig. 431 for the de~ 
tail. Negative 2"s or s""s 
point upward. 

From Fig. 431a 

I\{yz')=HJc 
,\ from symmetry 

Z% V z')=2H„k. 
From Fig. 4316 we have 



Fig. 431. 

and from Fig. 431c 



II(yzf f ')=H Q kr 
[The same pole distance H is taken in all these construc- 
tions] .-. I\{yz")=H {) {k^k T ). 

If, then, H {k\-\-k v ) = 2R k condition (9) is satisfied by the 
z' r "s. If not, the true pole distance for the special equil. 
polygon of Fig. 430 will be 

k Y -\-k c 



H- 



2k 



H 



With this pole distance and a pole in the horizontal through 
n'" (Fig. 430) the force diagram may be completed for the 
required special polygon ; and this latter may be con- 
structed as follows : Beginning at the point E, in vm, 
through it draw a segment II to the proper ray of the force 
diagram. In our present figure (430) this " proper ray " 
would be the ray joining the pole with the point of meet- 
ing of P 2 and P 3 on the load-line. Having this one seg- 



470 MECHANICS OF ENGINEEKING. 

ment of the special polygon the others are added in an 
obvious manner, and thus the whole polygon completed. 
It should pass through L, but not and B. 

For another loading a different special equil. polygon 
would result, and in each case we may obtain the thrust, 
shear, and moment of stress couple for any cross-section of 
the rib, by § 367. To the stresses computed from these, 
should be added (algebraically) those occasioned by change 
of temperature and by shortening of the rib as occasioned 
by the thrusts along the rib. These " temperature 
stresses," and stresses due to rib-shortening, will be con- 
sidered in a subsequent paragraph. They have no exist- 
ence for an arch-rib of three hinges. 

[Note. — If the moment of inertia of the rib is variable 

z' z" 

we put — for z' and — for z" in equation (6), (7), (8), and 



(9), n having the meaning given in the Note in § 379 a, 
which see ; and proceed accordingly]. 

381a. Exaggeration of Vertical Dimensions of Both Space and 
Force Diagrams. — In case, as often happens, the axis of the 
given rib is quite a flat curve, it is more accurate (for find- 
ing M) to proceed as follows : 

After drawing the curve in its true proportions and pass- 
ng a vertical through the middle of each of the equal 
4s's, compute the ordinate (y) of each of these middle points 
from the equation of the curve, and multiply each y by 
four (say). These quadruple ordinates are then laid off 
from the span upward, each in its proper vertical. Also 
multiply each load, of the given loading, by four, and then 
with these quadruple loads and quadruple ordinates, and 
the upper extremities of the latter as points in an exagge- 
rated rib-curve, proceed to construct a special equilibrium 
polygon, and the corresponding force diagram by the 
proper method ( for Class B, C, or Z>, as the case may be) 
for this exaggerated rib-curve. 

The moment, Hz, thus found for any section of the ex.- 



ARCH-RIBS. 471 

aggerated rib-curve, is to be divided by four to obtain the 
moment in the real rib, in the same vertical line. To find 
the thrust and shear, however, for sections of the real rib, 
besides employing tangents and normals of the real rib wo 
must draw, and use, another force diagram, obtained from 
the one already drawn (for the exaggerated rib) by re- 
ducing its vertical dimensions (only), in the ratio of four 
to one. [Of course, any other convenient number besides 
four, may be adopted throughout.] 

382. Stress Diagrams.— Take an arch-rib of Class D, § 378, 
i.e., of fixed ends,.and suppose that for a given loading (in- 
cluding its own weight) the special ^^^ thrust 
equil. polygon and its force diagram 

have been drawn [§ 381]. It is re- *!||§§ffj " "couple 

quired to indicate graphically the 
variation of the three stress-elements 
for any section of the rib, viz., the 
thrust, shear, and mom. of stress- 
couple. i~ is constant. If at any Exo.4as 
point m of the rib a section is made, then the stresses in 
that section are classified into three sets (Fig. 432). (See 
§§ 295 and 367) and from § 367 eq. (3) we see that the ver- 
tical intercepts between the rib and the special equil. 
polygon being proportional to the products Hz or 
moments of the stress-couples in the corresponding sec- 
tions form a moment diagram, on inspection of which we 

can trace the change in this moment, Hz = ^- , and 

e 

hence the variation of the stress per square inch, p 2 , (as 
due to stress couple alone) in the outermost fibre of any 
section (tension or compression) at distance e from the 
gravity axis of the section), from section to section along 
the rib. 

By drawing through lines On' and Of parallel re- 
spectively to the tangent and normal at any point m of the 
rib axis [see Fig. 433] and projecting upon them, in turn, 
the proper ray (R 3 in Fig. 433) (see eqs. 1 and 2 of § 367) 




tj 



MECHANICS OF ENGINEERING. 



472 



we obtain the values of the thrust and shear for the sec- 
tion at m. When found in this way for a number of points 
along the rib their values may be laid off as vertical lines 
from a horizontal axis, in the verticals containing the re- 
spective points, and thus a thrust diagram and a shear dia- 
gram may be formed, as constructed in Fig. 433. Notice 
that where the moment is a maximum or minimum the 
shear changes sign (compare § 240), either gradually or 




Fig. 433. 



suddenly, according as the max. or mm. occurs between 
two loads or in passing a load ; see m', e. g. 

Also it is evident, from the geometrical relations involv- 
ed, that at those points of the rib where the tangent-line 
is parallel to the " proper ray " of the force diagram, the 
thrust is a maximum (a local maximum) the moment (of 



ARCH-RIBS. 473 

stress couple) is either a maximum or a minimum and the 
shear is zero. 

.From the moment, Hz = -?-!_, p 2 — — t- 

e 1 

may be computed. From the thrust = Fp u p { =- — ==— , (F 

F 

— area of cross-section) may be computed. Hence the 
greatest compression per sq. inch (pi~\-p 2 ) may be found in 
each section. A separate stress-diagram might be con- 
structed for this quantity (pi+p 2 ). Its max. value (after 
adding the stress due to change of temperature, or to rib- 
shortening, for ribs of less than three hinges), wherever it 
occurs in the rib, must be made safe by proper designing 
of the rib. The maximum shear J m can be used as in §256 
to determine thickness of web, if the section is I-shaped, 
or box-shaped. See § 295. 

383. Temperature Stresses, — In an ordinary bridge truss 
and straight horizontal girders, free to expand or contract 
longitudinally, and in Classes A and B of § 378 of arch- 
ribs, there are no stresses induced by change of tempera- 
ture ; for the form of the beam or truss is under no 
constraint from the manner of support ; but with the arch- 
rib of two hinges (hinged ends, Class C) and of fixed ends 
(Class D) having immovable piers which constrain the dis- 
tance between the two ends to remain the same at all tem- 
peratures, stresses called " temperature stresses " are in- 
duced in the rib whenever the temperature, t, is not the 
same as that, t , when the rib was put in place. These 
may be determined, as follows, as if they were the only 
ones, and then combined, algebraically, with those due to 
the loading. 



384. Temperature Stresses in the Arch-Rib of Hinged Ends. — 
(Class C, § 378.) Fig. 434. Let E and 7 be constant, with 



474 MECHANICS OE ENGINEERING. 

._ at o t h ,, other postulates as in § 379. 

v"**^^^T^\ J^et £ = temperature of 

// y=z I s \ erection, and I = any other 

D y / H H \ b temperature ; also let I = 

£^jo~" "y length of span = 05 (in- 

" f^"^4. variable) and yj— co-efficient 

of linear expansion of the 
material of the curved beam or rib (see § 199). At tempera- 
ture t there must be a horizontal reaction H at each hinge 
to prevent expansion into the form O'B (dotted curve), 
which is the form natural to the rib for temperature t and 
without constraint. We may .*. consider the actual form 
OB as having resulted from the unstrained form O'B by 
displacing 0' to 0, ie., producing a horizontal displace- 
ment O'O =l(t-t )y. 

But O'O = Ax (see §§ 373 and 374) ; (N.B. B's, tangent 
has moved, but this does not affect Ax, if the axis X is 
horizontal, as here, coinciding with the span ;) and the 
ordinate y of any point m of the rib is identical with its 
z or intercept between it and the spec, equil. polygon, 
which here consists of one segment only, viz. : OB. Its 
force diagram consists of a single ray X n' ; see Fig. 434. 
Now (§ 373) 

Ax = JL CMyds ; and M=Hz = in this case, By 

. 7 (f _ t n r _ H /»*, . j hence for graphics, and 
• • H* M T i—fiijy as > \ equal As% we have 

Ell(t-t ) V =HAsI*f .... (1) 

From eq. (1) we determine H, having divided the rib-curve 
into from twelve to twenty equal parts each called As . 

For instance, for wrought iron, t and t , being expressed 
in Fahrenheit degrees, yj = 0.0000066. If E is expressed 
in lbs. per square inch, all linear quantities should be in 
inches and 2f will be obtained in pounds. 

I*y 2 may be obtained by § 375, or may be computed. H 
being known, we find the moment of stress-couple = Hy, 



ARCH-RIBS. 475 

-at any section, while the thrust and shear at that section 
are the projections of H, i.e., of x n' upon the tangent and 
normal. The stresses due to these may then be determined 
in any section, as already so frequently explained, and 
then combined with those due to loading. 

385. Temperature Stresses in the Arch-Ribs with Fixed Ends- 
See Fig. 435. (Same postulates as to symmetry, E and 1 
Constant, etc., as in § 380.) t and t have the same meaning 
as in § 384. 

Here, as before, we _ , 

consider the rib to m ' H 

have reached its ac- v^^TV'^^^v 

tual form under tern- , L // ■ -U T - * \ v § . N 

perature t by having / / // I \ \ \ \ 

had its span forcibly j | // \\ ) \ 

shortened from the _i^V/fo l_ 1 ^rV I 

length natural to ° "" ' l ~^~" *~ * 

temp, t, viz. : O'B' ', 

to the actual length OB, which the immovable piers compel 
it to assume. But here, since the tangents at and B are. 
to be the same in direction under constraint as before, the two 
forces H, representing the action of the piers on the rib, 
must be considered as acting on imaginary rigid prolonga- 
tions at an unknown distance d above the span. To find 
H and d we need two equations. 

From § 373 we have, since M=Hz—H{y—d), 

Ax, i.e., WO+SW, i.e., l(t-t )r],= m f(y-d)yds . (2) 

or, graphically, with equal As's 

EIl{t-t )r i =HJs\z* tf-dZ B o y~\ ... (3) 

Also, since there has been no change in the angle between 
end-tangents, we must have, from § 374, 

iif) Ms=0; L *';g3^*=o;i.e. l fjy- d y ds=0 



47G MECHANICS OF ENGINEERING. 

or for graphics, with equal yJ.s's, I*y = nd . . . (4): 

in which n denotes the number of Js's. From (4) we 
determine d f and then from (3) can compute H. Drawing 
the horizontal F G, it is the special equilibrium polygon 
(of but one segment) and the moment of the stress-couple 
at any section = Hz, while the thrust and shear are the 
projections of H= y n' on the tangent and normal respect- 
ively of any point m of rib. 

For example, in one span, of 550 feet, of the St. Louis 
Bridge, having a rise of 55 feet and fixed at the ends, the 
force H of Fig. 435 is = 108 tons, when the temperature is 
80° Fahr. higher than the temp, of erection, and the en- 
forced span is 3^ inches shorter than the span natural to 
that higher temperature. Evidently, :.f the actual temp- 
erature t is lower than that t , of erection, iTnmst act in a 
direction opposite to that of Figs. 435 and 434, and the 
" thrust " in any section will be negative, i.e., a pull. 

386. Stresses Due to Rib-Shortening —In § 369, Fig. 407, the 
shortening of the element AE to a length A'E, due to the 
uniformly distributed thrust, p x F } was neglected as pro- 
ducing indirectly a change of curvature and form in the 
rib axis ; but such will be the case if the rib has less than 
three hinges. This change in the length of the different 
portions of the rib curve, may be treated as if it were due 
to a change of temperature. For example, from § 199 we 
see that a thrust of 50 tons coming upon a sectional area 
of F = 10 sq. inches in an iron rib, whose material has a 
modulus of elasticity = E — 30,000,000 lbs. per sq. inch, 
and a coefficient of expansion rj = .0000066 per degree 
Fahrenheit, produces a shortening equal to that due to a 
fall of temperature (t Q —t) derived as follows : (See § 199) 
(units, inch and pound) 

(t -t)= P - 100 ' 0Q0 =50° 

Vo ; FEri 10 x 30,000,000 x. 0000066 

Fahrenheit. 

Practically, then, since most metal arch bridges of 

classes C and D are rather flat in curvature, and the thrusts 



AKCH-BIBS. 477 

due to ordinary modes of loading do not vary more than 20 
or 30 per cent, from each other along the rib, an imagin- 
ary fall of temperature corresponding to an average thrust 
in any case of loading may be made the basis of a con- 
struction similar to that in § 384 or § 385 (according as the 
ends are hinged, or fixed) from which new thrusts, shears, 
and stress-couple moments, may be derived to be combin- 
8d with those previously obtained for loading and for 
change of temperature. 

387. Resume — It is now seen how the stresses per square 
inch, both shearing and compression (or tension) may be 
obtained in all parts of any section of a solid arch-rib or 
curved beam of the kinds described, by combining the re- 
sults due to the three separate causes, viz.: the load, 
change of temperature, and rib-shortening caused by the 
thrusts due to the load (the latter agencies, however, com- 
ing into consideration only in classes C and D, see § 378). 
That is, in any cross-section, the stress in the outer fibre 
is, [letting T h ', T h '\ T h '", denote the thrusts due to the 
three causes, respectively, above mentioned ; (Hz)', (Hz)", 
(Hz)'" , the moments] 

i[(^)'±(^)"±(&r] . . . (i) 

i.e., lbs. per sq. inch compression (if those units are used). 
The double signs provide for the cases 
where the stresses in the outer fibre, due 
to a single agency, may be tensile. Fig. 
436 shows the meaning of e (the same 
used heretofore) /is the moment of in- 
ertia of the section about the gravity 
axis (horizontal) G. F = area of cross- 
section. [e { = e ; cross section symmet- 
rical about G\ For a given loading we 
may find the maximum stress in a given rib, or design the 
rib so that this maximum stress shall be safe for the ma- 
terial employed. Similarly, the resultant shear (total, not 



Tj±T h "+T h '" e 




478 



MECHANICS OF ENGINEERING. 



per sq. inch) = J' ± J" ± d'" is obtained for any section 
to compute a proper thickness of web, spacing of rivets, 
etc. 



388 The Arch-Truss, or braced arch. An open-work 
truss, if of homogeneous design from end to end, may be 
treated as a beam of constant section and constant moment 
of inertia, and if curved, like the St. Louis Bridge and the 
Coblenz Bridge (see § 378, Class D), may be treated as an 
arch-rib.* The moment of inertia may be taken as 



2 2^-iV 



2; 



where F Y is the sectional area of one of the pieces (I to the 
curved axis midway between them, Fig. 437, and h = dis- 
tance between them. 



l^v^O 




Fig. 438. 



Fig. 437. 



Treating this curved axis as an arch-rib, in the usual 
way (see preceding articles), we obtain the spec, equil. pol. 
and its force diagram for given loading. Any plane ~| to 
the rib-axis, where it crosses the middle m of a " web- 
member," cuts three pieces, A, B and 0, the total com- 



*The St Louis Bridge is not strictly of constant moment of inertia, being somewhat 
strengthened near each pier 



ARCH-RIBS. 479 

pressions (or tensions) in which are thus found : For the 
point ra, of rib-axis, there is a certain moment = Hz, a 
thrust = T h , and a shear = J, obtained as previously ex- 
plained. We may then write P sin ft = J (1) 

and thus determine whether P is a tension or compres- 
sion ; then putting P'+P" ± P cos fi = T h . . . . . 2 
(in which P is taken with a plus sign if a compression, and 
minus if tension); and 

{P r -P")\=Hz (3) 

A 

we compute F and P", which are assumed to be both com- 
pressions here, /? is the angle between the web member 
and the tangent to rib-axis at ra, the middle of the piece. 
See Fig. 406, as an explanation of the method just 
adopted. 



HORIZONTAL, STRAIGHT GIRDERS. 

389. Ends Free to Turn. — This corresponds to an arch- 
rib with hinged ends, but it must be understood that there 
is no hindrance to horizontal motion. (Fig. 439.) In 




Fig 439. 



treating a straight beam, slightly bent under vertical forces 
only (as in this case with no horizontal constraint), as a 



480 MECHANICS OF ENGINEERING. 

particular case of an arch-rib, it is evident that since the 
pole distance must be zero, the special equil. polygon will 
have all its segments vertical, and the corresponding force 
diagram reduces to a single vertical line (the load line). 
The first and last segments must pass through A and B 
(points of no moment) respectively, but being vertical will 
not intersect P x and P 2 ; i.e., the remainder of the special 
equilibrium polygon lies at an infinite distance above the 
span AB. Hence the actual spec, equil. pol. is useless. 

However, knowing that the shear, J, and the moment 
M (of stress couple) are the only quantities pertaining to 
any section ra (Fig. 439) which we wish to determine (since 
there is no thrust along the beam), and knowing that an 
imaginary force H ', applied horizontally at each end of the 
beam, would have no influence in determining the shear 
and moment at m as due to the new system of forces, we 
may therefore obtain the shears and moments graphically 
from this new system (viz.: the loads P l9 etc., the vertical 
reactions V and V n , and the two equal and opposite i?'"s). 
[Evidently, since H'' has no moment about the neutral 
axis (or gravity axis here), of ra, the moment at ra will be 
unaffected by it ; and since H" has no component ~\ to the 
beam at ra, the shear at ra is the same in the new system 
of forces, as in the old, before the introduction of the 
ZTs.] 

Hence, lay off the load-line 1 . . 2. . 3, Fig. 439, and con- 
struct an equil. polyg. which shall pass through A and B 
and have any convenient arbitrary H" (force) as a pole 
distance. This is done by first determining n' on the load- 
line, using the auxiliary polygon A'a'B\ to a pole 0' (arbi- 
trary), and drawing O'n' || to A'B'. Taking r on a hori- 
zontal through n\ making 0''n'=H", we complete the 
force diagram, and equil. pol. AaB. Then, z being the ver- 
tical intercept between ra and the equil. polygon, we have: 
Moment at m~M=K"z {oy—H'z' also), and shear at ra, or 
J,=2 . . n', i.e., = projection of the proper ray B 2 , or 
0" . . 2, upon the vertical through ra. Similarly we ob- 
tain M and J at any other section for the given load. (See 



ARCH-RIBS ; SPECIAL CASE ; STRAIGHT. 



481 



§§ 329, 337 and 367). The moment of inertia need not be 
constant in this case. 

390. Straight Horizontal Prismatic Girder of Fixed Ends at Same 
Level. — No horizontal constraint, hence no thrust. I con- 
stant. Ends at same level, with end-tangents horizontal. 
We may consider the whole beam free (cutting close to the 
walls) putting in the unknown upward shears J and J Qi 




and the two stress couples of unknown moments M and 
M n at these end sections. Also, as in § 388, an arbitrary 
H" horizontal and in line of beam at each extremity. Now 
(See Fig. 33) the couple at and the force H" are equiv- 
alent to a single horizontal H" at an unknown vertical dis- 
tance c below 0; similarly at the right hand end. The 
special polygon FG is to be determined for this new sys- 
tem, since the moment and shear will be the same at any 
section under this new system as under the real system. 
The conditions for determining it are as follows : Since 
the end-tangents are fixed, 2MAs =0 .*. Z%zds=0 and since 



482 MECHANICS OF ENGINEERING. 

O's displacement relatively to i?'s tangent is zero we have 
IMxAs =0 ,\ ZH"zxAs=Q .*. SxzAa =0. See § 374. Hence 
for Equal ^s's, l\z)=0 and I{xz)—0. Now for any pole 0'" 
draw an equil. pol. F'"G'" and in it (by § 377; see Note) 
locate v?"m'" so as to make Z(z"') = and i'(ccg / ")=0. 
Draw verticals through the intersections E f " and L"\ to 
determine E and L on the beam, these are the points of 
inflection (i.e., of zero moment), and are points in the re- 
quired special polygon FG. 

Draw 0"'n" || to v'"m f " to &x n". Take a pole 0" on 
the horizontal through n", making 0"n"=H" (arbitrary), 
draw the force diagram 0" 1234 and a corresponding 
equilibrium polygon beginning at E. It should cut X, 
and will fulfil the two requirments 2*(z)=0 and 2 , J(a?«)=0, 
with reference to the axis of the beam O'B'. The moment of 
the stress-couple at any section m will be M=H"z, and the 
shear J = the projection of the " proper ray " of the force 
diagram 0" . . 1, 2, etc., upon the vertical (not in the trial 
diagram 0"\ . 1, 2, etc.). As far as the moment is concern- 
ed the trial polygon F'" G'" will serve as well as the special 
polygon FG ; i.e., M=H'"z'" as well as H'% H"' being the 
pole-distance of 0'" ; but for the shear we must use the 
rays of the final and not the trial diagram. 

The peculiarity of this treatment of straight beams, 
considered as a particular case of curved beams, consists 
in the substitution of an imaginary system of forces in- 
volving the two equal and opposite, and arbitrary H's, for 
the real system in which there is no horizontal force and 
consequently no " special equilibrium polygon," and thus 
determining all that is desired, i.e., the moment and shear 
at any section. 

In the polygon FG the student will recognize the " mo- 
ment-diagram " of the problems in Chaps. Ill and IV. 

He will also see why the shear is proportional to the 

slope — — of the moment curve in those chapters. For 
ax 

example, the " slope " of the second segment of the poly- 
gon FG, that segment being || to 0" 2, is 



ARCH-RIBS; SPECIAL CASE; STRAIGHT. 483 



tang, of angle 20'V'=2w"-*-0'W'=sliear -- H" 

and similarly for any other segment ; i.e., the tangent of 
the inclination of the " moment curve," or line, is propor- 
tional to the shear. 

It is also interesting to notice with the present problem 
of a straight beam, that in the conditions 

I(zJs)=0 and l(zjds)x=0 9 

for locating the polygon FG> each ds is ~] to its 2, and 
that consequently each zAs is the area of a small vertical 
strip of area between the beam and the polygon, and 
[zAs)x is the " moment" of this strip of area, about 0' the 
origin of x. Hence these conditions imply ; first, that the 
area EWL between the polygon and the axis of the beam 
on one side is equal to that (0'FE-\-LB' G) on the other 
Side, and, secondly, that the centre of gravity of EWL lies 
in the same vertical as that of O'FE and LB 1 G combined. 
Another way of stating the same thing is that, if we join 
FG, the area of the trapezoid FO'B' G is equal to that of the 
figure FEWLG, and their centres of gravity lie in the same 
vertical. A corresponding statement may be made (if we join 
F"'G"') for the trapezoid F" , v'"m'"G'" and figure 
F" r E"'W'"L" , G"'< 



484 



MECHANICS OF ENGINEERING. 



CHAPTEE XII. 



GRAPHICS OF CONTINUOUS GIRDERS. 



[Mainly due to Prof. Mohr, of Aix-la-Chapelle ] 

391. The Elastic Curve of a Horizontal Loaded Beam, Homoge« 
neous and Originally Straight and Prismatic, is an Equilibrium 
Polygon, whose "load-line " is vertical \ j 

and consists of the successive products 
Mdx [treated as if they were loads A 
each applied through the middle of 
its proper dx], and tvhose "pole dis- 
tance " is EL Fig 441 (exaggerated). 

Let AO and OG be any two con- 
secutive equal elements of a very 
flat elastic curve (as above described). 
Prolong AO to cut NO. Then from 
§ 231, eq. (7), we have 




Fig. 442. 



dx 



(Mdx) 
~ET 



(1) 



where d 2 y=DC ;&nd, hence, if a triangle (Fig. 442) O'D'C, 
be formed with O'D' || to OD, O'C || to 00, and D'C ver- 
tical, while its (horizontal) altitude O'n is made equal, by 
scale, to EI of the beam, then from the similarity of the 
triangle ODG and O'D'C and the proportion in eq. (1) we 
see that D'C must represent the product Mdx on the same 
scale by which O'n represents EL 31 is fche moment of 



CONTINUOUS GIRDER BY GRAPHICS. 485 

the stress-couple at the section whose neutral axis is pro- 
jected in 0. 

Similarly, if M r is the moment of the stress-couple at 
0, and we draw O'F', \\ to CF, G'F' must represent M'dx 
(on sam® scale). It is therefore apparent that the line 
AOGF bears to the figure O'B'G'F' the same relation 
which an equilibrium polygon (for vertical forces) does to 
its force diagram, the " loads " of the force-diagram being 
the successive values of Mdx laid off to scale, while its 
"pole-distance " is inlaid off on the same scale. [As if 
Mdx and M'dx were loads suspended at and C respec- 
tively.] 

Practically, since any actual elastic curve is very flat, 
and since a change of pole-distance will change all verti- 
cal dimensions of the equilibrium polygon in an inverse 
equal ratio, we may exaggerate the vertical dimensions of 
the elastic curve by choosing a pole distance smaller than 
EI in any convenient ratio, n. Any deflection in the elas- 
tic curve thus obtained will be greater than its true value 
in the same ratio n. 

Graphically, in order to draw exaggerated elastic curves 
according to this principle, we obtain approximate results 
by dividing the length of the beam into a number of equal 
z/cc's, draw verticals through the middles of the Ax's as 
" force-verticals, "and lay off as a " load-line " to any con- 
venient scale the corresponding values of M Ax in their 
proper order. 

The quality of the product MAx is evidently (length) 2 x 
force, and with the foot and pound as units such a product 
may be called so many (sq. ft.) (lbs.). It will be noticed 
that these products (MAx) are proportional to, and maybe 
represented by, the areas of the corresponding vertical 
strips of the " moment-diagram " proper to the case in 
hand, These strips together make up the " moment-area" 
as it may be called, lying between the moment curve and 
its horizontal axis (which is the axis of the beam itself, 
according to §§ 389 and 390). 



486 MECHANICS OF ENGINEERING. 

392. Mohr's Theorem. — The principle of the previous 
paragraph may therefore be enunciated as follows : That 
just as the moment curve (of a straight prismatic horizontal 
beam) between two consecutive supports is an equilibrium poly- 
gon for the loading between those supports, so also is the elastic 
curve itself an equilibrium polygon for the " moment-area " con- 
sidered as a loading. 

In dealing with the moment-curve of a single span the 
pole distance is arbitrary (§§ 389 and 390), but the position 
of the pole relatively to the load line in other respects, and 
the location of the moment-curve (equil.-pol.) relatively to 
the beam (considered to be still straight for this purpose), 
depend on whether the beam simply rests on the two sup- 
ports, without projecting beyond ; or is built in, and at 
what angles ; or as with a continuous girder, on the inclina- 
tion of the tangent-lines at the supports, as influenced by 
the presence of loads on all the spans, and on whether all 
supports are on the same level or not. 

For example, in § 389, for a single span, the ends of 
beam being simply supported without overhanging, the 
pole 0" must be on a horizontal through n' , and the mo- 
ment curve must pass through the extremities A and B of 
the beam, thus giving a " moment-area " lying entirely on 
one side of the beam (or axis from which the moment or- 
dinates, z, are to be measured) ; whereas, in § 390, also a 
single span, where the ends of the beam are built in hori- 
zontally and at the same level, the pole must be taken on 
the horizontal through n" , and the moment-curve FEWLG 
must intersect the beam in the points E and L (E, L, and 
n" being found as prescribed in that problem), and thus 
lies partly above and partly below the beam. It will be 
necessary, later, to distinguish the upper and lower parts 
of the moment-area as positive and negative. 

In drawing the equilibrium polygon which constitutes 
the actual elastic curve, however, and hence making use of 
the successive small vertical strips of the moment-area, 
(when found) as if they were loads, to form a load-line ac- 
cording to a convenient scale, the pole distance is not ar- 



CONTINUOUS GIKDER BY GRAPHICS. 



487 



bitrary but must be = EI on the same scale. Still, since 
for convenience we must always greatly exaggerate the 
vertical scale of the elastic curve, we may make the pole 
distance = El-^-n and thus obtain an elastic curve whose 
Vertical dimensions are n times as large as those of the real 
curve ; while the position of the pole will depend in the 
direction of the tangent lines at the extremities of the span. 
An example will now be given. 

393. Example of an Elastic Curve (Beam Prismatic) Drawn as 
an Equilibrium Polygon Supporting the Moment-Area as Loading. 

■ — Let the beam be simply supported at its extremities (at 
the same level), and bear a single eccentric load P, Fig # 
443, its own weight being neglected. The moment-area 




consists of a triangle ACB [see first part of § 260, or use 

the graphic method of § 389, thus utilizing a force diagram 

012.], its altitude being the moment represented by the 

PI I 
ordinate CD and having a value = 12 . Hence the total 

c 

moment-area = ]/ 2 base AB x mom. CD 

i. e ., = *JA x y 2 1 = # fu, 



488 MECHANICS OF ENGINEERING. 

Divide AB into (say) eight equal dx's (eight are rather 
few in practice ; sixteen or twenty would be better) and 
draw a vertical through the middle of each. Note the 
portion of each of these vertical intercepts between the 
axis of the beam and the moment-curve ACB. The pro- 
ducts M Axiov the different subdivisions are proportional 
to these intercepts, since all the Ax's are equal, and are the 
respective moment-areas of the Ax's. 

Treating these products as if they were loads, we lay off 
the corresponding intercepts (or their halves, or quarters, 
or other convenient fractional part or multiple), from E 
downwards to form a vertical " load-line," beginning with 
fche left-hand intercept and continuing in proper order. 

As to what scale this implies, we determine by dividing 
the total moment-area thus laid off, viz. : y 2 Plil 2 , say in 
(sq. in.) (lbs.), by the length of EF in inches, thus obtain- 
ing the number of (sq. in.) (lbs.) which each linear inch of 
paper represents. 

On this scale the number of inches of paper required to 
represent the EI oi the beam is so enormous, that in its 
stead we use the nih. portion, n being an arbitrary abstract 
number of such magnitude as to make EI -4- n a con- 
venient pole-distance, TS. 

The proper position of the pole 0' on the vertical TW, 
is fixed by the fact that the elastic curve, beginning at A, 
must terminate in B, at the same level as A. Hence, 
assuming any trial pole as 0", and drawing rays in the 
usual manner (except that, as henceforth, the pole is taken 
on the right of the " load-line," instead of on the left, so 
as to make the resulting equilibrium polygon correspond 
in direction of curvature to the actual elastic curve), we 
draw the corresponding equilibrium polygon A"B". De- 
termining n' by drawing through 0" a line II to the right 
line A"0", we draw a horizontal through n' to intersect 
TW in 0' ', the required pole. 

"With f as pole a new equilibrium polygon begun at A' 
will terminate in B' and its vertical ordinates will be n 
times as great as those of corresponding points on the 



CONTINUOUS GERDER BY GEAPHICS. 489 

actual elastic curve AB. The same relation holds between 
the tangents of the angle of inclination to the horizontal 
at corresponding points (i.e., those in same vertical)of the 
two curves. 

394. Numerical Case of Foregoing Example. — With the inch 
and pound as units, let P = 120 lbs., ^ = 40 in., l 2 = 80 in. 
while the prismatic beam is of timber having a modulus 
of elasticity E = 2,000,000 lbs. per sq. inch, and is rectan- 
gular in section, being 2 in. wide and 4 in. high, so that 
(its width being placed horizontally) the moment of iner- 
tia of the section is / = 1 /„bh* = y i2 x2x64 = 10^ bi- 
quadratic inches (§ 90.) Bequired the maximum deflection. 
Adopting 1:20 as the scale for distances (i.e., one linear 
inch of paper to twenty inches of actual distance) we make 
the horizontal AB 6 in. long, Fig. 443, and AD 2 in., tak- 
ing a point G at convenience in the vertical through D, 
and joining AG and GB, thus determining the moment- 
diagram for this case. [As to what pole distance, H, is 
implied in this selection of (7, is immaterial in this simple 
case of a single load ; hence we do not draw the corre- 
sponding force- diagram at all.] We divide AB into eight 
equal parts and draw a vertical through the middle of 
each. The intercepts, in these verticals, between AB and 
the broken line AGB we lay off from E toward F as pre- 
scribed in § 393. (By taking DG small enough the line EF 
will not be inconveniently long.) 

Suppose this length EF measures 6.4 inches on the 
paper (as in the actual draft by the writer). Since it rep- 
resents a moment-area of 

^PZ 1 Z 8 =^xl20x40x80=192,000 (sq. in.) (lbs.), the scale 
of our "moment-area-diagram," as we may call it, must be 
192,000-7-6.4=30,000 (sq. in.) (lbs.) per linear inch of paper. 

Now #1=21,333,333 (sq. in.) (lbs.), which on the above 
scale would be represented by 711 linear inches of paper. 
With n = 100, however, we lay off ST=EI+n=lM inches 
of paper as a pole distance, and with a trial pole G" in 



490 MECHANICS OF ENGINEERING. 

the vertical TW draw the trial equilibrium polygon or 
elastic curve A"B", and with it determine n', then the final 
polygon A'B' as already prescribed. In A'B' we find the 
greatest ordinate, NK, to measure 0.88 inches of paper, 
which represents an actual distance of 0.88 x 20 = 17.6 
inches But the vertical dimensions of the exaggerated 
elastic curve A'B' are n=100 times as great as those of the 
actual, hence the actual max. deflection is d=17.6-t-n=0.176 
in. [This maximum deflection could also be obtained from 
the oblique polygon A"B" whose vertical dimensions are 
equal to those of A'B'. By the formula of § 235 we ob- 
tain d=0.174 inches.] 

395, Direction of End-Tangents of Elastic Curve in the Foregoing 
Problem. — As an illustration bearing on subsequent work 
let us suppose that the only result required in § 394 is 
tan « , i.e., the tangent of the angle B'A'T', which the 
tangent -line B'T' to the elastic curve at the extremity A', 
Fig. 343, makes with the horizontal line A'B', (tan. a is 
called the "slope" at A.) Let B'S' be the tangent-line at 
B'. These two " end-tangents " are parallel respectively to 
EO' and FO', and intersect at some point B. Now since 
A' KB' is an equilibrium polygon sustaining an imaginary 
set of loads represented by the successive vertical strips of 
the moment-area ACB, the intersection R must lie in the 
vertical containing the centre of gravity, U, of that mo- 
ment-area [§ 336 1 . 

Hence, if the vertical containing U is known in advance, 
or, as in the present case, is easily constructed without 
making the strip-subdivision of § 394, we may determine 
the end-tangents very briefly by considering the whole 
moment-area, M.A., (considered as a load) applied in the 
vertical through Z7, as follows : 

Since ACB is a triangle, we find U by bisecting AB in 
X, joining CX, and making XTJ = % XC, and then draw a 
vertical through U. Laying off EF=6A inches [so as to 
represent a moment-area of 192,000 (sq. in.) (lbs.) on a 
scale of 30,000 (sq. in.) (lbs.) per linear inch of paper], 



CONTINUOUS GIRDER BY GRAPHICS. 491 

and making ST=7.11 inches as before, we assume a trial 
pole 0" on TW, draw the two rays 0"E and 0"F> construct 
the corresponding trial polygon of two segments A"R"B", 
for the purpose of finding n' , With a pole 0' on TW and 
on a level with n' we draw the two rays O'E and O'F, and 
the corresponding segments A'R, and RB'. (B f should be 
on a level with A', as a check.) These two segments are 
the end-tangents required. 
We have, therefore, 



tan a = B'T' + A'B' = B"T' + A'B' 

In the present, numerical problem we find B'T' to mea- 
sure 3 in. of paper, i.e., 60 in. of actual distance for the 
exagg. elastic curve, and therefore 0.60 in. in the real elas- 
tic curve (with n = 100) 

, 0.60 in. A Ar w K 

.-. tan. a = __ — = 0.005 
120 m. 

It is now evident that the position and direction of the 
end-tangents of the elastic curve lying betiueen any two sup- 
ports are independent of the mode of distribution of the 
moment-area so long as the amount of that moment-area and 
the position of its centre of gravity remain unchanged. This 
relation is to be of great service. 

396. Re-Arrangement of the Moment- Area. — As another illus- 
tration conducing to clearness in later constructions, let 
us determine by still another method the end-tangents of 
the beam of §§ 394 and 395. See Fig. 444. As already 
seen, their location is independent of the arrangement of 
the moment-area between. Let us re-arrange this moment- 
area, viz., the triangle ACB, in the following manner : 

By drawing AX parallel to BC, and prolonging BC to V 
in the vertical through A, we may consider the original 
moment-area ACB to be compounded of the positive mom.- 
area VBXA, a parallelogram, with its gravity-vertical 
passing through D, the middle of the span ; of the negative 
mom. -area VCA, a triangle whose gravity-vertical passes 



492 



MECHANICS OF ENGINEERING. 



through B l making AD^ % AD ; and of another negative 
mom. -area, the triangle ABX, whose gravity -vertical passes 
through D 3 at one-third the span from B. That is, the 
(ideal) positive load ACB is the resultant of the positive 
load (M.A.) 2 and the two negative loads (or upward pulls) 
(M.A.) i and (M.A.) 3 , and may therefore be replaced by them 
without affecting the location of the end-tangents, at A 



H-lOin-. 




and B, of the elastic curve AB. These three moment- 
areas are represented by arrows, properly directed, in the 
figure, but must not be confused with the actual loads on 
the beam (of which, here, there is but one, viz., P). 

From the given shapes and dimensions, since ACB = 
192,000 (sq. in.) (lbs.), we easily derive by geometrical 
principles : 

(31A.) 2 = +576,000 (sq. in.) (lbs.) 
(M.A.\= - 96,000 
(M.A.) S = -288,000 

Hence, with a pole distance EI -f- n = 7.11 in. as before, 
and a " moment-load-line " formed of 1'2' = (M.A.) l9 (on 
scale of 30,000 (sq. in. lbs.) to one inch) 2'3' = (M.A.} 2 , and 
3'4' = (M.A.) Zi first with a trial pole 0", construct the trial 



CONTINUOUS GIRDER BY GRAPHICS. 493 

polygon A"B", and find n' in usual way (§ 337) ; then take 
& pole 0' on the horizontal through n' and the vertical 
TW, and draw the new polygon A' 123 B\ It should 
pass through B' on a level with A\ and A'\ and B'3 are the 
required end-tangents (of the exagg. elastic curve). 

[Note — If B' were not at the same level as A', but 
(say) 0.40 in. below it, B' should be placed at ra, a distance 

' X = 2 inches (on the paper) below its present posi- 

tion, (since the distance scale is 1:20 and n = 100, in this 
case) and the " abutment -line " of final polygon would be 
A'm\ 

Of course, this special re-arrangement of the moment- 
area is quite superfluous in the present problem of a dis- 
continuous girder (not built in), but considerations of this 
kind will be found indispensable with the successive spans 
of a continuous gi»der. 

397. Positive and Negative Moment- Areas in Each Span of a 
Continuous Girder (Prismatic).— In the foregoing problem of a 
discontinuous girder (covering one span only) not built in at 
ihe ends (otherwise it would be classed among continuous 
girders), the moment-curve, or equilibrium polygon of 
arbitrary H, is easily found by § 389 without the aid of the 
elastic curve, and the end moments are both zero ; (i.e., the 
moment-curve meets the beam in the end- verticals) but in 
each span of a continuous girder the end-moments are not 
zero (necessarily), and the points in the end-verticals where 
the moment-curve must terminate (for an assumed H) can 
not be found without the use of the elastic curve (or of 
some of its tangents) of the whole beam, dependent, as it is, 
upon the loading on all the spans, and the heights of the 
supports. 

Let Fig. 445 show, in general, any one span of a pris- 
matic continuous girder {prismatic ; hence I is constant), 
between two consecutive supports A and B . P^ P 2 , etc., 
are the loads on the span. 

[If the displacement of A n relatively to the end-tangent 
.at B , and the angle between the end-tangents (of elastic 



494 



MECHANICS OF ENGINEERING. 



curve) were known, the. moment-curve or equilibrium 
polygon FWG, {AB being the axis of beam) might be 
found by a process similar to that in § 390, but the elastic 
curves in successive spans are so inter-dependent that the 
above elements can not be found directly.] 




Fig. 445. 

We now suppose, for the sake of discussion, that the 
whole girder has been investigated (by a process to be 
presented) for the given loads, spans, positions of supports, 
etc., and then the moment-curve FEWLG found, with the 
corresponding force-diagram, for the span in the figure 
and some arbitrary H. The horizontal line AB represents 
the axis of the beam (for this purpose considered straight 
and horizontal) as an axis from which to measure the 
moment ordinates. 

Thus, the moment (of the stress-couple) at A is = H X 
AF; at B t Hx BG ; at E andX, zero (points of inflection). 

Now, according to the usual conceptions of analytical 
geometry, we may consider the portion FWL, of the mo- 
ment-area, above AB as positive, and those below, AEF 
and LB G, as negative ; but since not one of these three 



CONTIGUOUS GIRDER BY GRAPHICS. 495 

areas, nor the position of its gravity-vertical, is known in 
advance, since they are not independent of the other 
spans, a more advantageous re-arrangement of the moment- 
area may be made thus : 

Join FG and FB, and we may consider the original 
moment-area replaced by the following three component 
areas : the positive moment-area FEWLGF (shaded by ver- 
tical lines) ; the negative triangular moment-area AFB ; 
and the negative triangular moment-area BFG (the nega- 
tive moment-areas being shaded by horizontal lines). (In 
subsequent paragraphs, by positive and negative moment- 
areas will be implied those just mentioned.) 

These three moment-areas, treated as loads, each applied 
in its own gravity -vertical, and considered in any order, 
may be used instead of the real distributed moment- area, 
as far as determining, or dealing with, the end-tangents of the 
elastic-curve at A 'and B Q is concerned (§ 396), and the fol- 
lowing advantages will have been gained : 

(1.) The amount of the positive moment-area, (M.A.) 2 in 
Fig. 445, (depending on the area lying between the polygon 
FEW G and the abutment-line F G of the latter), and the 
position of its gravity -vertical, are independent of other spans, 
and can be easily found in advance, since this moment-area 
and gravity -vertical are the same as if the part of the beam 
covering this span were discontinuous and simply rested on the 
supports A and B , as in § 389. 

(2.) The gravity-vertical of the left-hand negative mo- 
ment-area, (M.A.\, is always one-third the span from the 
left end-vertical, A A'; that of the other, (M.A.) 3 , an equal 
distance from the right end-vertical, B B'. 

(3.) The two (right and left) negative moment-areas 
are triangular, each having the whole span I for its alti- 
tude, and for its base the intercept AF {ox BG) on which 
the end-moment depends. Hence, if the amounts of thesa 
negative moment-areas have been found in any span, we 
may compute the values which AF and B G must have for 
a given H, and thus determine the terminal points F and 
G of the moment-curve of that span (for that value of H\ 



496 MECHANICS OF ENGINEERING. 

For example, if (M.A.\ has been found (by a process not 
yet given) to be 160,000 (sq. in.) (lbs.) while AB = I = 160 
in., then the moment M A which Misrepresents, is computed 
from the relation 

(M.A.\ = y 2 ~AB x M A 

or,I A= W°,»in,lbs, 
160 

If Zfhas been chosen = 100 lbs. we put Hx AF = 2000 
and obtain AF .= 20 inches of actual distance, so that with 
a scale of 1:20 for distances AF would be one linear inch 
of paper. (Of course, in computing BG the same value of 
H must be used.) With H= 100 lbs., then, and F and G 
as known points of the equilibrium polygon FEW G, it 
is easily drawn by the principles of § 341. 

We thus notice that the amounts of the two negative 
moment-areas are the only elements affected by the con- 
tinuity of the girder, in this re-arrangement of the actual 
moment-areas. 

In the lower part of Fig. 445 A' and B' represent the 
extremities of the (exagg.) elastic curve, the vertical dis- 
tance B'B"\ of B' from the horizontal through A' (in case 
the two supports A and B Q are not at same level, as we 
here suppose for illustration) being laid off in accordance 
with the principles of the note in § 396. 

Note. — It is now evident that if the "false polygon" (as 
it will be called) A'VZ&B' has been obtained (and means for 
doing this will be given later) in which the first and last 
segments are the end tangents of the (exagg.) elastic curve, 
and which bears the same relation to the three moment 
areas just mentioned, as that illustrated in Fig. 444, we 
may proceed further to determine the amounts of (M.A.) { 
and (M.A.) ?t as follows, by completing the moment-area dia- 
gram : 

Having laid off the known (M.A.) 2 (or positive moment- 
area)^^', and ST=EI+n, a line parallel to 12 drawn 
through 2', determines the pole 0', through which paral- 



CONTINUOUS GIRDER BY GRAPHICS. 



497 



lels to A'l and B'3 will fix 1' and 4' on the vertical SV, 
and thus determine l'2'=(ilO.)i and 3'4'=(Jf.^.) 3 . Their 
numerical values are then computed in accordance with 
the scale of the moment-area diagram. 

The polygon A'123B' will be called the "false polygon " 
of the span in question, its end-segments being the end- 
tangents of the elastic curve. 

398. Values of the Positive Moment-Area in Special Cases.— 
For several special cases these are easily computed, and 
as an illustration, Fig. 446 shows a continuous girder, AF 



UJHll 



! w" ! 

'UUUlHi: 




Fig. 44 



of five spans, all six supports on a level, and the weight of 
the beam neglected. At the extremities A and F, as at 
the other supports, the beam is not built in, but simply 
touches each support in one point ; hence the moments at 
A and F are zero, i.e., the moment curve must pass through 
A and F, so that in the first span the left negative mo- 
ment-area, and in the fifth span the right negative mo- 
ment-area, are zero. The positive moment-areas are 
shaded. 

On the first span is placed a uniformly distributed load 
W over the whole span V. .*. the positive moment-area 
for that span is the same as in the case of Fig. 235 [see 
(1) § 397] and being represented by a parabolic segment 
whose area is two-thirds that of the circumscribing rec- 
tangle, its value is 



498 MECHANICS OF ENGINEERING. 

{M.A.y 2 =y 3 .y 8 wi'xi'= 1 / 12 wp . . (i) 

[see eq. (2) § 242], while its gravity- vertical bisects the 
span. 

The cnly load on the second span is a concentrated one, 
P", at distances ?/' andZ 2 ' from the extremities of the span; 
hence the positive moment-area is triangular and has a 
value 

(M.A.) 2 "=y 2 rw . . . (2) 

as in § 393. Its gravity vertical may easily be constructed 
as in Fig. 443 [see (1) § 397]. 

The third span carries no load ; hence its positive mo- 
ment-area is zero, and the actual moment-area is composed 
solely of the two triangular negative moment-areas GDH 
and DHI, the moment-curve consisting of the single 
straight line HI. 

The fourth span carries a uniform load W IV =w 1Y l lv , and 
.-. has a positive moment-area 

{M.A.)Y= l / l2 W l \l™y . . . (3) 

as in eq. (1), acting through the middle of the span (grav- 
ity-vertical). 

Since the fifth and last span carries no load, its positive 
moment-area is zero, the moment curve being the straight 
line JF, so that the actual moment-area is composed of 
the left-hand negative moment-area. 

At F it is noticeable that the reaction or pressure of the 
support must be from above downward to prevent the 
beam from leaving the point F; i.e., the beam must be 
" latched down," and the reaction is negative. 

If the beam were built in at A (or F) the moment at 
that section would not be zero, hence the left (or right) 
neg. moment-area would not be zero in that span, as in 
our present figure. But in such a case the tangent of the 
elastic curve would have Sb'lcnown direction at A (or F) and 
the problem would still be determinate as will beseem 



CONTINUOUS GIKDEE, BY GRAPHICS. 499 

A f . . . F' gives an approximate idea (exaggerated) of 
the form of the elastic curve of the entire girder. A change 
in the loading on any span would affect the form of this 
curve throughout its whole length as well as of all the 
moment curves. 

Note. — It is important to remark that any two of the 
triangular negative moment-areas which haVe a common 
base (hence lying in adjacent spans) are proportional to 
their altitude i.e., to the lengths of the spans in which they 
occur ; thus the neg. mom.-areas (Fig. 446) GCH&n&DCIl 
have a common base CH, 

. (3i.A.y = r_ 

' ' {M.A.y r v ; 

(The notation explains itself ; see figure.) It also follows, 
that the resultant of these two neg. moment-areas (if re- 
quired in any construction ; see § 400) acts in a vertical 
ivhich divides the horizontal distance between their gravity ver- 
ticals in the inverse ratio of the spans to which they belong 
[§ 21 and eq. (4) above]. 

Hence, since this horizontal distance is /il ff -\-y^l"' their 
resultant must act in a vertical Y r/r , whose distance from 
the gravity-vertical of GCH is %l'", and from that of 

CHD, y 3 i. 

399. Amount and Gravity- Vertical of the Positive Moment 
Area of One Span as Due to Any Loading. — Since we can not 
deal directly with a continuous load by graphics, but must 
subdivide it into a number of detached loads sufficiently 
numerous to give a close approximation, let us suppose 
that this has already been done if necessary, and that P x , 
P 2 , etc., are the detached loads resting in the span AB in 
question ; see Fig. 447. 

Since [by (1), § 397] the positive moment-area is the 
same as the total moment-area would be if this portion of 
the beam simply rested on the extremities of the span, not 
extending beyond them, we may use the construction in § 
389 for finding it, remembering that in that paragraph the 



500 



MECHANICS OF ENGINEERING. 



oblique polygon in the lower part of Fig. 439 will serve 
as well as the (upper) one whose abutment -line is the 
beam itself, as far as moments are concerned. 

Hence, Fig. 447, lay off the load-line LL' 9 take any pole 
0, with any convenient pole-distance H, and draw the 
equilibrium polygon FWG. After joining FG, FWGF 
will be the positive moment-area required. 

To find its gravity vertical, divide the span AB, or FG\ 
into from ten to twenty equal parts (each = As) and draw a 




Pig. 447. 

vertical through the middle of each. The lengths z lf z 2> 
etc., on these verticals, intercepted in the moment -area, 
are proportional to the corresponding strips of moment- 
area, each of width =ds, and of an amount =HzAs. 

Form a load line, SK, of the successive a's, and with 
any pole 0', draw the equilibrium polygon A'B' (for the 
z-verticals). The intersection, B, of the extreme segments, 
is a point in the required gravity-vertical (§ 336). 

The amovfnt of the moment-area is (M.A.\-=2\_HzAs~] 

=KAs. Z(z)=H.4s(z l +z 2 +Zs+ . . . ) 



CONTINUOUS GIRDER BY GRAPHICS. 501 

For example, with the span =Z=120 in., subdivided into 
twelve equal As's, we have As -10 inches (of actual distance). 
If H=4 inches of paper and SK=I(z)=10.2 inches of paper, 
the force-scale being 80 lbs. to the inch, and the distance- 
scale 15 inches to the inch (1:15), we have 

(M.A.) 2 = [4 x 80] x 10 x [10.2x15] =489600. (sq. in.)(lbs.) 

400. Construction of the "False-Polygons" For All the Spans 
of a Given (Prismatic) Continuous Girder, Under Given Loading, 
and With Given Heights of Supports. — [See note in § 397 for 
meaning of " false-polygon ".] Let us suppose that the 
given girder covers three unequal spans, Fig. 448, with 
supports at unequal heights, and that both extremities A 
and D are built-in, or " fixed," horizontally. To clear the 
ground for the present construction, we suppose that, from 
the given loading in each span, the positive moment-area 
of each span has been obtained in numerical form [so 
many (sq. in.) (lbs.) or (sq. in.) (tons)] and its gravity-ver- 
tical determined by § 398 or § 399 ; that the horizontal 
distances (i.e., the spans I', I", and V" and the distance be- 
tween the above gravity-verticals and the supports) have 
been laid off on some convenient scale ; that EI has been 
computed from the material and shape of section of the 
girder and expressed in the same units as the above mo- 
ment-areas ; that a convenient value for n has been se- 
lected (since El-^n is to be the pole distance of all the 
moment-area-diagrams), and that the vertical distances of 
B, C, and D, from the horizontal through A, have been 
laid off accordingly (see note in § 396). 

In the figure (448) verticals are drawn through the 
points of support ; also verticals dividing each span into 
thirds, since the unknown negative moment-areas (sub- 
scripts 1 and 3) act in the latter (§ 397) ; and the gravity- 
verticals of the known positive moment-areas. The ver- 
ticals Y\ Y' f , V", and V"', are to be constructed later. 

The problem may now be stated as follows : 

Given the -positions of the supports, the value of EI and n f 



502 



MECHANICS OF EKGLNEEKING. 



the fact that the girder is fixed horizontally at A and D, the 
heights of supports, the loca Hon of the gravity -verticals of all 
the positive and negative moment-areas, and the amounts of the 
positive moment-areas ; it is required to find graphically the 
"false-polygon " in each span. 

The " false-polygons," viz.: A 123 B for the first span (on 
the left), B 123 G for the second, etc., are drawn in the figure 



• 1 ff A vll (I J; rflj ti 

(MlAXr/ i!\ I t J Ml ! K M Wi 




Fig. 448. 

for the purpose of discussing their properties at the out- 
set. Since SB and Bl are both tangent to the elastic curve 
at B, they form a single straight line ; similarly (71 is but 
the prolongation of 3(7. Also Al and 3D must be hori- 
zontal since the beam is built in horizontally at its ex- 
tremities A and D. 

That is, the three false polygons form a continuous 
equilibrium polygon A . . . D, in equilibrium under the 
" loads " 

(M.A.y, (M.A.y, (M.A.y, etc., 

so that we might use a single mom. -area-diagram in con- 
nection with it, but for convenience the latter will be 



COXTIXFOUS GIRDER BY GRAPHICS. 503 

drawn in portions, one under each span, with a pole dis- 

tance = — . 
n 

Of this polygon A ... D, we have the two segments Al 
and 3D already drawn, and know that it passes through 
the points B and G ; we shall next determine by construc- 
tion other points (called " fixed points ") p ', p", p Q ", p"\ 
and Pq" in (the prolongations of) certain other segments. 

To find the "fixed point " p % where the segment 23 in the 
first span cuts the vertical V, the gravity -vertical of 
(M.A.y. The vertex p\ or 1, is already known, being the 
intersection of Al with V. Lay off 2' 3' = (M.A.) 2 ' which is 
known, and take a trial pole 0/with a pole-distance EI -j- n ; 
join 0/ 2' and 0/ 3'. Draw 12 || to 2'0 t to determine 2 
on the vertical (M. A.\\ then through 2 a line || to 0/3' to 
cut V'mp '. The unknown segment 23 must cut F*'in 
the same point ; since all positions of 0/ on the vertical 
T'TJ' will result in placing^/ in this same point, and one 
of these positions must be the real pole 0' (unknown). 
[This is easily proved in detail by two pairs of similar 
triangles]. 

To determine the "fixed-point " p", in the prolongation 

of segment 12 of second span. The prolongations 

of the segments 12 (of first span) and 12 (of second span) 

must meet in a point h in the (vertical) line of action of the 

resultant of (M. A.\' and (if. A.y (§ 338). Although the 

amounts of {M. A.y and (31. A.)" are unknown, still the 

vertical line of action of their resultant (by § 398, Note) is 

V 
known to be Y', a horizontal distance — to the right of 

o 

(M.A.) 3 ' ; hence Y f is easily drawn. Therefore, the unknown 
triangle £'31 has its three vertices on three known verti- 
cals, the side k'3 passes through the known point p ', and 
the side 13 through the known point B. Now by pro- 
longing Pq 2 (or any line through p Q ') to cut (M. A.y and 
Y' in 3 and & ', respectively, joining 3 B and prolonging 
this line to cut (M. A.\" in some point 1 , and then joining 
Jc ' lo, we have a triangle k '3 l of which we can make a 



504 MECHANICS OF ENGINEERING. 

statement precisely the same as that just made for &' 3 L 

But if two triangles [as k'31 and V 3 1 ] have their 
vertices on three parallel lines (or on three lines which 
meet in a point) the three intersections of their correspond- 
ing sides must lie on the same straight line [see reference 
to Chauvenet, § 378 a]. Of these intersections we have 
two, p ' and B ; hence the third must lie at the intersection 
of the line p ' B (prolonged) with Tcq 1 , and in this way the 
"fixed point" p", a point in ¥1 and .*. in the segment 12 
(of second span; prolonged, is found. Draw a vertical 
through it and call it V". 

The fixed point p " (in prolongation of segment 23 of sec- 
ond span) lies in the vertical V" and is found from p" and 
the known value of (M.A.) 2 " precisely &$p ' was found from 
p\ That is, we lay off vertically 2" 3" = (M.A.) 2 ", and join 
2" and 3" to t ", which is any point at distance EI — n 
to the right of 2"3". Through p" draw a line || to 2" t " 
to cut (M. A.) 2 " in 2 , then 2 p " || to 0" 3" to determine 
p" on the vertical V". 

The fixed points p' '" andjV" in the third span lie in the 
prolongations of the segments 12 and 23, respectively, of 
that span, p"' being found from the points p Q " and G and 
the verticals (M.A.\" t Y" 9 and (31. A.)/", in the same manner 
as p" was determined with similar data, while Po'" 9 in the 
same vertical V" as p'" 9 depends on (M. A.) 2 '" and its 
gravity vertical as already illustrated ; hence the detail 
need not be given ; see figure. 

In this way for any number of spans we proceed from 
span to span toward the right and determine the succes- 
sive fixed points, until the points p and p of the last span 
have been constructed, which are p r " and p ' f ' in our 
present problem. Since p"' is a point in the segment 23 
(prolonged) of the last span, we have only to join it with 
3 in that span, a point already known, and the segment 23 
is determined. Joining the intersection 2 with p'" we 
determine the next segment 21 and of coarse the vertex 1, 
which is then joined with G and prolonged to intersect 
M. A.) 3 " to fix the segment 103 and the point 3. Join 



CONTINUOUS GIRDER BY GRAPHICS. 505 

3 p ", and proceed in a similar manner toward the left, 
until the whole equilibrium polygon (or series of " false 
polygons ") is finally constructed ; the last step being the 
joining of 2 with p'. 

401. Treatment of Special Features of the Last Problem. — (1.) 
If the beam is simply supported at A, Fig. 448, instead of 
built-in, (M. A.)/ becomes zero, and the two segments A\ 
and 12 of that span form a single segment of unknown 
direction. Hence, the point A will take the place of p', 
and the vertical V A that of V. 

(2.) Similarly, if D, in the last span, is a simple support 
(beam not built in) (M. A.) z '" becomes zero, and the seg- 
ments 1)3 and 32 form a single segment of unknown 
direction, so that after p /r has been found, we join p '" 
and I) to determine the segment D2 ; i.e., in this last span, 
D takes the place of 3 of the previous article. 

(3.) If the first span carries no load (M. A.) 2 ' is zero, and 
the segments 12 and 23 will form a single segment 23. 
Hence if the beam is built in at A, pj will coincide with 
the known point p' (i.e., 1), while if A is a simple support 
p and p coincide with A, since, then, (M A.){ is zero and 
A 123 is a single segment. 

(4.) If the last span is unloaded (third span in Fig. 448), 
(31. A.) 2 '" is zero, 123 becomes a single segment, and hence 
p"' will coincide with p" f ; so that after p fn has been con- 
structed it is to be directly joined to 3, if the beam is 
built in at I), and will thus determine the segment 13 ; or 
to D, if D is a simple support, (for then (M. A.) z "' is zero 
and the three segments 12, 23, and 3 D form a single seg- 
ment.) 

(5.) If an intermediate span is unloaded (say the second 
span, Fig. 448) the positive mom.-area, (M. A.) 2 "> is zero, 
123 becomes a straight line, i.e. a single segment, and 
therefore p Q " coincides with p" ; hence, when p" has been 
found we proceed as if it were p". 

402. To Find the Negative Mom. -Areas, the Mom. -Curves, Shears, 
and Reactions of the Supports. — (1.) Having constructed the 



o06 MECHANICS OF ENGINEERING. 

false polygons according to the last two articles, the nega- 
tive moment-areas of each span are then to be found by the 
note in § 397, Fig. 445, and expressed in numerical form. 

[If the positive mom.-area of the span is zero the points 
2' and 3' will coincide, Fig. 445, and in the case mentioned 
in (3), (or (4)), of § 401, if A (or D) were a simple support, 
in Fig. 448, the mom. -area-diagram of Fig. 445 would have 
but two rays.] 

(2.) The moments at the supports (or " end-moments " of 
the respective spans) depending, as they do, directly on 
the negative mom. -areas, can now be computed as illustrated 
in (3) § 397. The fact that each " end-moment " may be 
obtained from two negative mom. -areas, separately, one in 
each adjacent span (except, of course at the extremities of 
the girder) forms a check on the accuracy of the work. 
The two values should agree within one or two per cent. 

(3.) The " moment-curve " of each span or equilibrium 
polygon formed from a force-diagram whose load-line con- 
sists of the actual loads on the span laid off in proper 
order, can now be drawn, a convenient value for H having 
been selected (the same Hfor all the spans, that the moment- 
carves of successive spans may form a continuous line for 
the whole girder); since we may easily compute the proper 
moment ordinate at each support to represent the actual 
moment, then, for the H adopted, by (3) § 397. The 
moment-curve of each span, since we know its two extreme 
points and its pole-distance H, is then constructed by 
§ 341. 

(4.) The shear. Since the last construction involves 
drawing the special force-diagram for each span, with a 
ray corresponding to each part of the span between two 
consecutive loads, the shear at any section of the beam is 
easily found as being the length of the vertical projection 
of the " proper ray," interpreted by the force-scale of the 
force-diagram, as in §§ 389 and 390. With the shears as 



CONTINUOUS GIRDER BY GRAPHICS. 



507 




Fig. 449. 

librium, i.e., IY = (§ 36), we have 



ordinates a shear -diagram may 
now be constructed, if desired, for 
each span. The directions of the 
shears should be carefully noted. 
(5.) Reactions of supports. Let 
us consider " free " the small por- 
tion of the girder, at each point 
of support, included between two 
sections, one close to the support 
on each side, Fig. 449. Suppose 
it is the support C, and call the 
reaction, or pressure at that sup- 
port B c . Then, for vertical equi- 



ne— ^CR+eA: 



CL 



(5) 



and, in general, the reaction at a support equals the 
(algebraic) sum of the two shears, one close to the support 
ou the right, the other on the left. The meaning of the 
subscripts is evident. In applying this rule, however, a 
free body like that in Fig. 449 should always be drawn, or 
conceived ; for the two shears are not always in the same 
direction ; hence the phrase " algebraic sum." 

At a terminal support, as A or F, Fig. 446, if the beam 
is not built in, the reaction is simply equal to the shear 
(since the beam does not overhang) just as in §§ 241 and 
243. Fig. 446 presents the peculiarity that the reaction 
of the support F is negative, (as compared with R G in Fig. 
449); i.e., the support at F must be placed above the beam 
to prevent its rising (this might also be the case at (7, or 
D, in Fig. 446, for certain relations between the loads). 



403. Numerical Example of Preceding Methods. — As illustrat- 
ing the constructions just given, it is required to investi- 
gate the case of a rolled wrought-iron " I-beam," [a 15- 
inch heavy beam of the N. J. Steel and Iron Co.,] extend- 
ing over four supports at the same level, covering three 



508 MECHANICS OF ENGJ^EEKOGr. 

spans of 16, 20, and 14 feet respectively, and bearing a 
single load in each of the extreme spans, but a uniform 
load over the entire central span. As indicated thus : 

7 ft. 9 ft. 20 ft. 8 ft. 6 ft. 
A I B G I D 



30 tons 40 tons 32 tons 

[This is a practical case where W" is the weight of a 
brick wall, and P' and P"' are loads transmitted by col- 
umns from the upper floors of the building ; A and D are 
simple supports, and the weight of the girder is neglected.] 

The beam has a moment of inertia / == 707 biquad. 
inches, and the modulus of elasticity of the iron is E = 
25,000,000 lbs. per sq. in., = 12,500 tons per sq. in. 

Although with a prismatic continuous girder under 
given loading, with supports at the same level, it may easily 
be shown that the moments, shears, and reactions, to be 
obtained graphically, are the same for all values of i, so 
long as the elastic limit is not surpassed, still, on account 
of the necessity of its use in other problems (supports 
not on a leveD, we shall proceed as if the value of / were 
essential in this one. 

Selecting the inch and ton as units for numerical work, 
we have 

EI= 12500 X 707. = 8,837,500 (sq. in.) (tons) while the 
respective positive mom. -areas, from eqs. (2) and (3) of 
§ 398, are : 

(M. A.y =}4 X 30 X 84 x 108 = 136,080 (sq. in.) (tons) 
(M. A.\" = y i2 x 40 x 240 2 = 192,000 " " 
(M. A.) 2 '" = y 2 x 32 x 96 x72 = 110,592 " " 

Adopting a scale of 60,000 (sq. in.) (tons) to the linear inch 
of paper, for mom.-area diagrams, we have for the above 
mom.-areas 2.27 in., 3.20 in., and 1.84 in., respectively, on 
the paper, for use in Fig. 448. 



CONTINUOUS GIKDEE, BY GRAPHICS. 509 

Having laid off the three spans on a scale of 60 inches 
to the inch of paper, with A, B, C, and D in the same hori- 
zontal line, we find by the construction of Fig. 443, that the 
gravity -vertical of (M. A.}/ lies 3.6 in. to the left of the 
middle in the first span, that of (M. A.) 2 '" 4.8 in. to the 
right of the middle of the third span ; while that of 
(M. A.) 2 ", of course, bisects the central span. Hence we 
draw these verticals ; and also those of the unknown neg- 
ative mom. -areas through the one-third points ; remember- 
ing [§ 401, (1) and (2)] that (M. A.)/ and (M. A.y are 
both zero in this case. 

Since EI = 8,837,500 (sq. in.) (tons), it would require 
147.29 in. to represent it, as pole-distance, on a scale of 
60,000 (sq. in.) (tons) to the inch ; hence let us take n = 50 
for the degree of (vertical) exaggeration of the false poly- 

gons, since the corresponding pole-distance — = 2.94 in. 

of paper is a convenient length for use with the values of 
(M. A.) 2 ', (M. A.) 2 ", etc., above given. 

Following the construction of Fig. 448, except that p' is 
at A, and p '" is to be joined to D (§ 401), (the student will 
do well to draft the problem for himself, using the pre- 
scribed scales,) and thus determining the false-polygons, 
we then construct and compute the neg. mom. -areas 
according to § 402 (1), and the note in § 397, obtaining the 
following results ; 

(M. A.y, 1.43 in. of pap., = 85,800 (sq. in.) (tons) 
(M. A.\", 1.77 " " " = 106,200 " " 
(M. A.y, 1.68 " " " = 100,800 " " 
(M. A.\'",l.n « " " = 70,200 " " 



The remaining results are best indicated by the aid of 
Fig. 450. Following the items of § 402, we find [(3) § 397] 
that the moment at B, using (M. A.y, is 

M 2X106,200 = g85 inch . toM 
240 m. 



510 MECHANICS OF ENGINEERING. 

[or, using (M. A.y, \M B = 2x g 00 =893 in. tons.] 

Similarly, M G = 2x ^ 800 = 840 in. tons, 
[or, using (M. A.\'" 9 M c = 835.7 in. tons.] 
Hence, taking means, we have, finally, 

M A =0 ; 1T B =889 in. tons ; ^=837.8 ; i!f D =0. 

Fig. 450 shows the actual mom.-areas and shear-dia« 
grams, which are now to be constructed. 




Selecting a value H — 20 tons for the pole-distance of 
the successive force-diagrams, (the scale of distances being 
5 ft. (60 in.) to the inch we have [(3) § 397] 

20 x ~BG = M B = 889 in. -tons .-. BG = 44.4 in. of actual 
distance, or 0.74 in. of paper ; also 20 x GK = M Q = 837.8 
in. -tons .*. GK = 41.89 in., or 0.698 in. of paper. 

Having thus found (rand K, and divided BG into ten 
equal parts, applying four tons in the middle of each, we 
construct by § 341 an equilibrium polygon which shall 
pass through G and iT and have 20 tons as a pole-distance. 
(We take a force-scale of 10 tons to the inch.) It will 
form a (succession of short tangents to a) parabola, and is 
the moment curve for span BG, Similarly, for the single* 



CONTINUOUS GIRDER BY GRAPHICS. 511 

loads P' and P'" in the other two spans, we draw the 
equilibrium polygons AN'G and KZ"'D, for the same II 
as before, and passing through A and G, and K and B, re- 
spectively. 

Scaling the moment -ordinates NN', QQ", and ZZ"\ 
reducing to actual distance and multiplying by H t we have 
for these local moment maxima, i!f N = 1008, M q = 336, and 
M z = 936, in. tons. 

Evidently the greatest moment is i!f N and .*. the stress 
in the outer fibre at N will be (§ 239) 

jgy=^g= 100 ^ x7 ^=10.0 tons per sq. inch which is much 

too large. If we employ a 20-inch heavy beam, with / = 
1650 biquad. in., the preceding moments will still be the 
same (supports all at same level) and we have 

1008x10 arka , 
P* = -, ggft = 6.06 tons per sq. m., 

or nearly 12,000 lbs. per sq. in., and is therefore safe 
(§183). 

If three discontinuous beams were to be used, the 20- 
inch size of beam (heavy) would be much too weak, in 
each of the three spans, as may be easily shown ; hence the 
economy of the continuous girder in such a case is readily per- 
ceived. It will be seen, however, that the cases of conti- 
nuity and of discontinuity do not differ so much in the 
shear-diagrams as in the moment curves. By scaling the 
vertical projection of the proper rays in the special force 
diagrams (as in §§ 389 and 390) we obtain the shear for 
any section on AN, as J AR (see Fig. 449 for notation) = 
12.3 tons ; on NB, J BL = 17.7 tons ; from B to G it varies 
uniformly from e/ BR = 20.3 tons, through zero at Q, to J CIj 
= 19.7 tons of opposite sign. Also, for CZ, J GR = 18.6 
tons ; and for ZD, J BJj = 13.4 tons. Hence, the reactions 
of the supports are as follows : 

B A =J AR =12.3 tons ; B B =J BL +J Br =38.0 tons. 
Po— '-^cr+^cl = 38.3 tons ; B D = J Dh =13A tons. 



512 MECHANICS OF ENGINEERING. 

[In the shear-diagram, the shear-ordinates are laid off 
below the axis when the shear points down, the "free body " 
extending to the right of the section considered, (as «7 CL in Fig. 
449) ; and above, when the shear points upward for the 
same position of the free body.] 

If we divide the max. shear, 20.3 tons by the area of the 
web, 13.75 sq. in., of the 20-inch heavy beam, (§256), we 
obtain 1.5 tons or 3000 lbs. per sq. in., which is < 4000 
(§ 183). Notice the points of inflection, i' , %", etc., where 
M is zero. 

Sufficient bearing surface should be provided at the 
supports. 

A swing-bridge offers an interesting case of a continu- 
ous girder, 

404. Continuous Girder of Variable Mom. of Inertia. — If 2" is 
variable and I denote the mom. of inertia of some con- 
venient standard section, then we may write I = I -s- m, 
when m denotes the number of times I contains I. In a 
non-prismatic beam, m is different for different sections 
but is easily found, and will be considered given at each 
section. 

In eq. (1) of § 391, then, we must put I Q -f- m in place of 
I and thus write 

d 2 y_ \mMdx\ ,fi 

dtf~ EI, l ' 

and (pursuing the same reasoning as there given) may 
therefore say that in a girder of variable section if each 
small vertical strip (Mdx) of the moment-area be multiplied by 
the value of m proper to that section, and these products (or "vir- 
tual mom. -area strips) considered as loads, the elastic curve is an 
equilibrium polygon for those loads with a pole distance = EI . 

In modifying § 400 for a girder of variable section, then, 
besides taking EI -f- n as pole distance, proceed as 
follows : 

Construct the positive mom.-area for each span accord- . 
ing to § 399 ; for each z of Fig. 447, substitute mz (each z 



CONTIGUOUS GIRDER BY GRAPHICS. 513 

having in general its own m), and thus obtain the " virtual 
positive mom. -area," and its gravity vertical. 

Similarly, there will be an unknown " virtual neg. mom.' 
area" not triangular, replacing each neg. mom.-area of 
§ 400. Though it is not triangular, each of its ordi- 
nates equals the corresponding ordinate of the unknown 
triangular neg. mom.-area multiplied by the proper m, and 
its gravity -vertical (ichich is independent of the amount of the 
unhioivn neg. mom.-area) is found in advance by the process 
of Fig. 447, using, for s's, a set of ordinates obtained thus: 
Draw any two straight lines AB and FB, Fig. 445, (for a 
left-hand trial'neg. mom.-area; or FB and GFiov a right- 
hand one) meeting in the end-vertical of the span, divide 
the span into ten or twenty equal spaces, draw a vertical 
through the middle of each, noting their intercepts between 
AB and FB. Add these intercepts and call the sum S. 
Multiply each intercept by the proper m, and with these 
new values as z's construct their gravity vertical as in Fig. 
447. Add these new intercepts, call the sum S vi and denote 
the quotient S -s- S Y by j3. 

We substitute the three verticals mentioned, therefore, 
for the mom.-area verticals of § 400, and the " virtual pos. 
mom.-area " for the pos. mom.-area, in each span ; pro- 
ceed in other respects to construct the "false polygons" 
according to § 400. Then the result of applying the con- 
struction in the note § 397 will be the "virtual neg. mom.- 
areas," each of which is to be multiplied by the proper 
/3 to obtain the -corresponding triangular neg. mom.-area, 
with which we then proceed, without further modifica- 
tions in the process, according to (2), (3), etc. of § 402. 

[The conception of these " virtual mom.-areas " is due 
to Prof. Eddy ; see p. 36 of his " Eesearches in Graphical 
Statics," referred to in the preface of this work.] 

405. Remarks. — It must be remembered that any unequal 
settling of the supports after the girder has been put in 
place, may cause considerable changes in the values of the 
moments, shears, etc., and thus cause the actual stresses to 
be quite different from those computed without taking 



514 MECHANICS OF ENGINEERING. 

into account a possible change in the heights of the sup- 
ports. See § 271. 

For example, if some of the supports are of masonry, 
while others are the upper extremities of high iron or 
steel columns, the fluctuations of length in the latter due 
to changes of temperature will produce results of the 
nature indicated above. 

If an open-work truss of homogeneous design from end 
to end (treated as a girder of constant moment of inertia, 
whose value may be formulated as in § 388,) is used as a 
continuous girder under moving loads, it will be subject 
to " reversal of stress " in some of its upper and lower hor- 
izontal members, i.e., the latter must be of a proper de- 
sign to sustain both tension and compression, (according 
to the position of the moving loads,) and this may disturb 
the assumption of homogeneity of design. Still, if / is 
variable, § 404 can be used ; but since the weight of the 
truss must be considered as part of the loading, several 
assumptions and approximations may be necessary before 
establishing satisfactory dimensions. 



INDEX. 



Page. 

Aberration of Light 90 

Absolute Velocity 89 

Abutment-Line 414 

Abutments of Arches 430, 485 

Acceleration, Angular 107 

Acceleration, Linear 49 

Acceleration, Nprmal 75, 77 

Acceleration of Gravity 51, 160, 179 

Acceleration, Tangential 74, 80 

Action and Reaction 1, 53 

Angular, or Rotary, Motion. . . 107 

Anomalies in Friction 192 

Anti- Derivative, see Preface 

and p. 253 

Anti-Resultant 402 

Anti-Stress-Resultant 411 

Apparent Weight 78, 79 

Arches Linear 386, 396 

Arches of Masonry 421, 437 

Arch-Ribs 438, 483 

Arch-Ribs, Classification of 458 

Arch-Rib of Three Hinges . 458,460 
Arch-Rib of Hinged Ends 440. 

458, 461 

Arch-Rib of Fixed Ends 439,459,465 
Arch Truss, or Braced Arch.. . 478 

Atwood's Machine 159 

Autographic Testing Machine. 240 
Beams, Rectangular, Compar- 
ative Strength and Stiffness 

272,273,277 

Belting, Pressure of 181 

Belting, slip of 182 

Bent Lever with Friction 173 

Boat-Rowing 160 

Bow's Notation '.'.407 

Box-Girder 275,292 

Braced Arch 438,478 

Brake, Prony 158 

Brakes, Railroad .. ... 190 

Bridge, Arch 430 

Bridge-Pier 141 

Bridge, Suspension '. ...... 46 

Bridge Truss, Warren 35 

Buckling of Web-Plates 383 

Built Beams, designing Sec- 
tions of 295 

Built Columns 378 

Burr, Prof., Citations from. 224, 229 
Butt-Joint 226 



Page.. 

Cantilevers 260, 276, 341 

Cantilever, Oblique 353, 356 

Catenary, Common 46 

Catenary Inverted 387 

Catenarv, Transformed 895 

Cast Iron 220, 279 

Cast Iron, Malleable 224 

Centre of Gravity, 18, 19, etc.. . .336 

Centre of Oscillation 119 

Centre of Percussion of Rod. ..Ill 

Centrifugal Action 125 

Centrifugal Force 77, 78 

Centripetal Force 77, 78, 79 

Centrobaric Method 24 

Cheval- Vapeur 13d 

Chrome Steel .. . . 224 

Circle as Elastic Curve 262 

Circular Arc as Linear Arch . . 391 

Closing Line 414 

Coblenz, Bridge at 459, 478 

Columns, Long 363 

Composition of Forces... 4, 8, 31, 38 
Compression of Short Blocks. . 218 

Concurrent Forces 6, 8, 397 

Cone of Friction 168 

Conical Pendulum 78 

Connecting-Rod 59, 69, 70 

Conservation of Energy 156 

Conservation of Momentum 66 

Continuous Girders, by Analysis 

320—332 

Continuous Girders, by 

Graphics 484—514 

Copying-Press 71 

Cords, Flexible 42 

Cords. Rigidity of 192 

Couples 27. 30 

Cover-Plates 226 

Crane 362 

Crank-Shaft. Strength of 314 

Creeping of Belts 186 

Crushing, Modulus of 219, 424 

Curvilinear Motion 72 

Dangerous Section 262, 332 

Dash-Pot 158 

Deck-Beam 275 

Deflections, (Flexure) 252—262, 342 
Derivatives, (Elastic curve).... 310 

Derived Quantities 2 

Deviating Force 77 



"xn 



IXDEX. 



Diagrams, Strain 209, 241 

Displacement of Point of Arch 

Rib 447 

Dove-Tail Joint 269 

Duchayla's Proof of the Parallelo- 
gram of Forces 4 

Dynamics, Definition 4 

Dynamics of a Rigid Body 105 

Dynamics, of Material point. ... 49 

Dynamometers 157, 158, 159 

Eddy, Prof., Graphic Methods, 

(See Preface) 426 

Elastic Curve a Circle . . . 262, 34a 
Elastic Curve an Equilibrium 

Polygon 484 

Elastic Curves. . 245, 252—262, 35Q 
Elastic Curves, the Four ^-De- 
rivatives of 310 

Elasticitv-Line 241 

Elasticity, Modulus of 203, 22? 

Elastic Limit 203 

Elevation of Outer Rail on Rail- 
road Curves. 78 

Ellipse of Inertia 94 

Ellipsoid of Inertia 104 

Elliptical Beam 340 

Elongation of Wrought-Iron 

Rod 207 

Energy 137 

Energy, Conservation of 156 

Energy, Kinetic... 137, 144, 147, 150 

Energy, Potential . . . 155 

Equations, Homogeneous 2 

Equator, Apparent Weight at the 78 

Equilibrium 4, 33, 39 

Equilibrium Polygon 401,450 

Equilibrium Polygon Through 

Three Points 418,419 

Equivalent Systems 7, 105, 145 

Exaggeration of Vertical Di- 
mensions in Arch-Ribs. . . . 470 

Examples in Flexure 280—284 

Examples in Shearing 231, 232 

Examples in Tension and 

Compression 222, 223 

Examples in Torsion 241—243 

Experiments of an English 

Railroad Commission 314 

Experiments of Hodgkinson 

207, 369 

Experiments of Prof. Lanza... 280 
Experiments on Building Stone 424 

Experiments on Columns 378 

Extrados , 421 

Euler's Formula for Columns. . . 364 

Factor of Safety 223 

Falling Bodies 51 

"Fa ise Polygons' 1 497, 501 

Fatigue of Metals 224 



" Fixed Points " 503 

Flexural Stiffness 250 

Flexure .244—386 

Flexure and Torsion Combined 314 
Flexure, Beams of Uniform 

Strength 335 

Flexure, Common Theory 244 

Flexure, Eccentric Load.. .256, 301 
Flexure, Elastic Curves in.. 245, 

251, 252—262 

Flexure, Examples in 280—284 

Flexure, Hydrostatic Pressure 308 

Flexure, Moving Loads 298 

Flexure, Non-Prismatic 

Beams 332, 335 

Flexure of Long Columns 363 

Flexure of PrismaticBeams 

Under Oblique Forces. . .347—362 

Flexure, Safe Loads in 262—284 

Flexure, Safe Stress in 279 

Flexure, Shearing Stress in 

284—295 

Flexure, Special Problems 

in 295—319 

Flexure, Strength in 249 

Flexure, the Elastic Forces 246 

Flexure, the "Moment" 249 

Flexure, the "Shear" 248 

Flexure, Uniform Load. . . 258, 267 

302, 305, 307, 324, 329, 340 

Flow of Solids 212 

Fly- Wheel 121, 151 

Fly- Wheel, Stresses in. .. .126, 127 

Force 1 

Force Diagram 400 

Force Polygons 397 

Forces, Concurrent 6, 8 

Forces, Distributed 197 

Forces, Non-Concurrent 6, 31 

Forces, Parallel 13 

Forces, Parallelogram of 4 

Forces, Varieties of 7 

Free Axes 129 

Free-Body Method, the 11, 105 

Friction 164—194, 422, 423 

Frictional Gearing 172 

Friction, Anomalies in 192 

Friction Axle 175 

Friction Brake 158 

Friction, Cone of 168 

Friction in Machinery 191 

Friction, Sliding 164—168 

Friction of Pivots 179 

Friction, Rolling 186 

Friction-Wheels : 177 

Friction, Work Spent in 149 

General Properties of Materials 

. 204 

Governor Ball 78 



INDEX. 



Xlll 



Graphic Representations of Uni- j 
formally Accelerated Motion. . 57 j 

Graphic Treatment of Arch 431 j 

Gravity, Acceleration of 51, 79, 160 

Gravity, Centre of 18, 336,453 

Gravity- Vertical 453 

Graphical Statics, Elements 

of 397—420 

Graphical Statics of Vertical 

Forces 412—420 

Guide Curve... 83 

Gyroscope 132 

Harmonic Motion 58, 81, 117 

Heat Energy 156 

Heaviness, Table, Etc 3 

Height Due to Velocity 52, 84 

Hodgkinson's Formulae for 

Columns '. 369 

Homogeneous Equations 2 

Hooke's La,w 201, 203, 207 

Hooks, Strength of 362 

Horizontal Straight Girders by 

Graphics 479—483 

Horse-Power 136, 239,242 

I-Beam 275, 292,295,337 

Ice-Boat, Speed of 90 

Impact 63 

Impact, Loss of Energy in. . . . 141 

Inclined Beam 1 359 

Inclined Plane 83, 135, 151, 166, 169 

Indicator 159 

Inertia 53 

Inertia of Piston-Rod 59 

Instantaneous Rotation, Axis 

of 112 

Internal Stress, General Prob- 
lem of 205 

Intrados 421 

Isochronal Axes 120 

Isotropes 204 

Kinetic Energy... 137,144, 147, 150 

Knot, Fixed 43 

Knot, Slip 43 

Lanza, Experiments of Prof. . . . 280 

Lateral Contraction 211, 229 

Lateral Security of Girders 280, 298 

Lever 18, 71 

Lever, Bent, With Friction 173, 174 
Linear Arches... 386—396, 417, 425 

Live Loads 298, 430 

Load-Line .♦. . 413 

Locomotive on Arch 430 

Locomotive on Girder 298 

Locomotive. Parallel-Rod of... 131 

Malleable Cast Iron 224 

Mass 2, 53 I 

Material Point 3 | 

Mechanical Equivalent of Heau 156 j 
Mechanics, Definition of 1 ' 



Mechanics, Divisions of 4 

Middle Third 423 

Modulus of Elasticity 203, 227 

Modulus of Resilience 213 

Modulus of Rupture (Flexure) 278 

Modulus of Tenacity 212 

Moduli of Compression 219* 

Mohr's Theorem 486 

Moment-Area 485 

Moment of a Force 14 

Moment-Diagram 263 

Moment of Flexure. . . 248, 348, 351 

Moment of Inertia 91 

Moment of Inertia by Graphics 454 
Moment of Inertia of Box- 
Girder 276 

Moment of Inertia of Built 

Beam 296 

Moment of Inertia of Built 

Column 379 

Moment of Inertia of Plane 

Figures 91—98, 249, 274 

Moment of Inertia of Rigid 

Bodies 98, 103 

Moment of Inertia of Truss 478 

Moment of Torsion 236 

Momentum 66 

Mortar 422 

Motion, Curvilinear 72 

Motion, Rectilinear 5ft 

Motion, Rotary 107 

Moving Loads (Flexure 298 

Naperian Base 183, 357, 387 

Naperian Logarithms 47 

Navier's Principle 422, 436 

Neutral Axis 245, 247, 347, 355 

Neutral Line (see Elastic Curve.) 

Newton's Laws .1, 53 

Non-Concurrent Forces in a 

Plane 31, 399 

Non-Concurrent Forces in Space 37 

Normal Acceleration 75, 76, 77 

Normal Stress 200 

Oblique Section of Rod in Ten- 
sion 200 

Parabola as Linear Arch 391 

Parabolic Cord 45 

Parabolic Working-Beam. 336, 344 

Parallel Forces 13 

Parallel-Rod of Locomotive. . 131 

Parallelogram of Forces 4 

Parallelogram of Motions 72 

Parallelogram of Velocities 72 

Pendulum, Compound 118 

Pendulum, Conical 78 

Pendulum, Cycloidal 80 

Pendulum, Simple Circular 81 

Phoenix Columns 378 

Pier Reactions 404 



XIV 



INDEX. 



Piers of Arches 430, 435 

Pile-Driving 140 

Pillars (see columns) 

" Pin-and-Square " Columns... 364 

Planet, Velocity of 82 

Polar Moment of Inertia, . . . 97, 238 

Pole (in Graphics) 401 

Pole-Distance 416, 417 

Practical Notes 223 

Potential Energy 155 

Power 134 

Power of Motors. . . . 153, 157, 158 
Power, Transmission of, by 

Belting 184 

Power, Transmission of by 

Shafts 238, 318 

Principal Axes 104, 129 

Projectile in Vacuo 83, 84—87 

Prony Friction Brake 158 

Pulley..., 43, 103 

Punching Rivet Holes 229 

Quantity, Kinds of 1 

Quantities, Derived 2 

Radius of Curvature, 75, 76, 

250, 353 

Radius of Gyration 91, 92, 115, 

.... 313,376 

Rankine's Formula for Col- 
umns 372, 375 

Rays of Force Diagram 401 

Reaction . , 1, 18, 36, 53, 404 

Reduced Load-Contour 429 

Regulation of Machines 153 

Relative and Absolute Veloci- 
ties 89 

Resilience.... 204, 213, 237, 251, 313 
Resultant of Parallel Forces, 13, 15 
Resultant of Two or More . 

Forces 4, 6 

Reversal of Stress 514 

Rigid Body 4 

Rigidity of Ropes 192 

Riveting for Built Beams. . . . 292 
Rivets and Riveted Plates. 225, 29-2 
Robinson, Prof. , Integration by 357 

Rod in Tension 198, 200 

Rolling Friction 186 

Rolling Motion 130 

Roof Truss 37, 405 

Rotary Motion 68, 107, 129 

Rotation and Translation Com- 
bined , -130, 150 

Rupture 202 

Safe Limit of Stress 202 

Safe Loads in Flexure . . . 262, 284 

St. Louis Bridge 459, 467, 478 

Schiele's ' 'Anti-Friction" Pivots 181 
Set, Permanent... 202, 209, 208, 241 
Shafts 233—239 



Shafts, Non-Circular 239 

Shear Diagram (Flexure) 265 

Shearing 225—232 

Shearing Distortion 227 

Shear, Distribution of in Flex- 
ure 287 

Shearing For.ces 7, 225 

Shearing Stress, 

7,200, 201, 225, 234, 284 

Shear, the First ^-Derivative 

of Moment (Flexure) 264 

Skidding 190 

Slip (of Oar, etc.) 161 

Slope (in Flexure) 253 

Soffit 421 

Spandrel 421 

Special Equilibrium Polygon 

409,424,440 

Specific Gravity 3 

Statics, Definition of 4 

Statics, Graphical 397—420 

Statics of a Material Point 8 

Statics of a Rigid Body 27 

Statics of Flexible Cords 42 

Steam Engine Problems... 59, 

61, 69, 70,121,131, 151 

Steam Hammer 138 

Stiffening of Web-Plates 383 

Stone, Strength of 221, 424 

Stress Diagrams for Arch-Ribs. 471 
Stresses Due to Rib Shorten- 
ing 476 

Strain Diagrams 209, 241 

Strains, two kinds only 196 

Stress 197, 198 

Stress and Strain, Relation 

Between 201 

Stress-Couple 253, 348 

Stress, Normal and Shearing... 200 

Strength of Materials 195 

Stretching of a Prism Under 

its Own Weight 215 

" Sudden " Application of a 

Load 214, 255 

Summation of Products by 

Graphics 451 

Suspension Bridge 46 

Table for Flexure 279 

Table for Shearing 228 

Tables for Tension and Com- 
pression 221 

Tackle 43 

Temperature Stresses 190, 199, 

222,473 

Tenacity, Moduli of 212 

Testing Machine, Autographic 240 
Theorem of Three Moments.... 332 

Thrust (in Flexure) 348, 350 

Torsion 233—243 



INDEX. 



XV 



Torsion, Angle of 233 

Torsion Balance 116 

Torsion, Helix Angle in 233 

Torsion, Moment of 236 

Torsional Resilience 237 

Torsional Stiffness 236 

Torsional Strength 235 

Tractrix, The 181 

Transformed Catenary 395 

Transmission of power by 

Belting 184 

Transmission of Power by Shafting 

238, 318 

Translation, Motion of 68, 

105,106, 133, 137 

Trussed Girders 381 

Uniformly Accelerated Motion 

.* 54, 107 

Uniform Motion 48, 107, 129 

Uniform Strength, Beams of.. 335 
Uniform Strength, Solid of, in 

Tension 216 



Units, Proper Use see § 6, p. 2 

Velocity, Absolute 89 

Velocity, Angular 107 

Velocity, Linear 49 

Velocity, Relative 89 

Virtual Moment 67 

Virtual Velocities 67 

Voussoir 386, 421 

Warren Bridge Truss 35 

Water. Jets of 87 

Web of I-Beam 274 

Web of I-Beam, Buckling of.. 383 

Web of I-Beam, Shear in 290 

Wedge, with Friction 171 

Weight 2,3,7,79 

Weight, Apparent. 78, 79 

Wind and Sail-Boat 89, 90 

Work 133, 134, etc. 

Work and Energy in Machines 

146,147, etc. 

Working-Beam 336, 344 

Working Strength 202 



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